Pandas 中文参考指南
Cookbook
这是一个存储实用的 pandas 配方的简洁、精确例子和链接的库。我们鼓励用户为该文档添加内容。
This is a repository for short and sweet examples and links for useful pandas recipes. We encourage users to add to this documentation.
添加有趣的链接和/或将内联示例添加到这一部分是一个绝佳的首次请求提取（Pull Request）。
Adding interesting links and/or inline examples to this section is a great First Pull Request.
在可能的情况下，已插入简化、浓缩、对新用户友好的内联示例以增强 Stack Overflow 和 GitHub 链接。很多链接都包含了比内联示例提供的更多的信息。
Simplified, condensed, new-user friendly, in-line examples have been inserted where possible to augment the Stack-Overflow and GitHub links. Many of the links contain expanded information, above what the in-line examples offer.
pandas (pd) 和 NumPy (np) 是仅有的两个缩写导入模块。其余的对于新用户来说保持显式导入。
pandas (pd) and NumPy (np) are the only two abbreviated imported modules. The rest are kept explicitly imported for newer users.
Idioms
以下是一些简洁的 pandas idioms
These are some neat pandas idioms
In [1]: df = pd.DataFrame(
...: {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]}
...: )
...:
In [2]: df
Out[2]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50if-then…
对一列执行 if-then
An if-then on one column
In [3]: df.loc[df.AAA >= 5, "BBB"] = -1
In [4]: df
Out[4]:
AAA BBB CCC
0 4 10 100
1 5 -1 50
2 6 -1 -30
3 7 -1 -50对 2 列执行 if-then 并赋值：
An if-then with assignment to 2 columns:
In [5]: df.loc[df.AAA >= 5, ["BBB", "CCC"]] = 555
In [6]: df
Out[6]:
AAA BBB CCC
0 4 10 100
1 5 555 555
2 6 555 555
3 7 555 555添加另一条包含不同逻辑的行，来执行 -else
Add another line with different logic, to do the -else
In [7]: df.loc[df.AAA < 5, ["BBB", "CCC"]] = 2000
In [8]: df
Out[8]:
AAA BBB CCC
0 4 2000 2000
1 5 555 555
2 6 555 555
3 7 555 555或者在设置掩码后使用 pandas
Or use pandas where after you’ve set up a mask
In [9]: df_mask = pd.DataFrame(
...: {"AAA": [True] * 4, "BBB": [False] * 4, "CCC": [True, False] * 2}
...: )
...:
In [10]: df.where(df_mask, -1000)
Out[10]:
AAA BBB CCC
0 4 -1000 2000
1 5 -1000 -1000
2 6 -1000 555
3 7 -1000 -1000In [11]: df = pd.DataFrame(
....: {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]}
....: )
....:
In [12]: df
Out[12]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [13]: df["logic"] = np.where(df["AAA"] > 5, "high", "low")
In [14]: df
Out[14]:
AAA BBB CCC logic
0 4 10 100 low
1 5 20 50 low
2 6 30 -30 high
3 7 40 -50 highSplitting
In [15]: df = pd.DataFrame(
....: {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]}
....: )
....:
In [16]: df
Out[16]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [17]: df[df.AAA <= 5]
Out[17]:
AAA BBB CCC
0 4 10 100
1 5 20 50
In [18]: df[df.AAA > 5]
Out[18]:
AAA BBB CCC
2 6 30 -30
3 7 40 -50Building criteria
In [19]: df = pd.DataFrame(
....: {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]}
....: )
....:
In [20]: df
Out[20]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50…并且（无赋值返回序列）
…and (without assignment returns a Series)
In [21]: df.loc[(df["BBB"] < 25) & (df["CCC"] >= -40), "AAA"]
Out[21]:
0 4
1 5
Name: AAA, dtype: int64…或者（无赋值返回序列）
…or (without assignment returns a Series)
In [22]: df.loc[(df["BBB"] > 25) | (df["CCC"] >= -40), "AAA"]
Out[22]:
0 4
1 5
2 6
3 7
Name: AAA, dtype: int64…或者（赋值时修改数据框。）
…or (with assignment modifies the DataFrame.)
In [23]: df.loc[(df["BBB"] > 25) | (df["CCC"] >= 75), "AAA"] = 999
In [24]: df
Out[24]:
AAA BBB CCC
0 999 10 100
1 5 20 50
2 999 30 -30
3 999 40 -50In [25]: df = pd.DataFrame(
....: {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]}
....: )
....:
In [26]: df
Out[26]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [27]: aValue = 43.0
In [28]: df.loc[(df.CCC - aValue).abs().argsort()]
Out[28]:
AAA BBB CCC
1 5 20 50
0 4 10 100
2 6 30 -30
3 7 40 -50In [29]: df = pd.DataFrame(
....: {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]}
....: )
....:
In [30]: df
Out[30]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [31]: Crit1 = df.AAA <= 5.5
In [32]: Crit2 = df.BBB == 10.0
In [33]: Crit3 = df.CCC > -40.0可以硬编码：
One could hard code:
In [34]: AllCrit = Crit1 & Crit2 & Crit3…或者用动态构建的条件的列表来实现
…Or it can be done with a list of dynamically built criteria
In [35]: import functools
In [36]: CritList = [Crit1, Crit2, Crit3]
In [37]: AllCrit = functools.reduce(lambda x, y: x & y, CritList)
In [38]: df[AllCrit]
Out[38]:
AAA BBB CCC
0 4 10 100Selection
Dataframes
indexing 文档。
The indexing docs.
In [39]: df = pd.DataFrame(
....: {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]}
....: )
....:
In [40]: df
Out[40]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [41]: df[(df.AAA <= 6) & (df.index.isin([0, 2, 4]))]
Out[41]:
AAA BBB CCC
0 4 10 100
2 6 30 -30使用 loc 进行面向标签的切片和 iloc 位置切片 GH 2904
Use loc for label-oriented slicing and iloc positional slicing GH 2904
In [42]: df = pd.DataFrame(
....: {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]},
....: index=["foo", "bar", "boo", "kar"],
....: )
....:有 2 个显式切片方法，第三个是一般情况
There are 2 explicit slicing methods, with a third general case
-
Positional-oriented (Python slicing style : exclusive of end)
-
Label-oriented (Non-Python slicing style : inclusive of end)
-
General (Either slicing style : depends on if the slice contains labels or positions)
In [43]: df.loc["bar":"kar"] # Label
Out[43]:
AAA BBB CCC
bar 5 20 50
boo 6 30 -30
kar 7 40 -50
# Generic
In [44]: df[0:3]
Out[44]:
AAA BBB CCC
foo 4 10 100
bar 5 20 50
boo 6 30 -30
In [45]: df["bar":"kar"]
Out[45]:
AAA BBB CCC
bar 5 20 50
boo 6 30 -30
kar 7 40 -50如果一个索引包含从一个非零开始的整数或非单位增量，则会出现二义性。
Ambiguity arises when an index consists of integers with a non-zero start or non-unit increment.
In [46]: data = {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]}
In [47]: df2 = pd.DataFrame(data=data, index=[1, 2, 3, 4]) # Note index starts at 1.
In [48]: df2.iloc[1:3] # Position-oriented
Out[48]:
AAA BBB CCC
2 5 20 50
3 6 30 -30
In [49]: df2.loc[1:3] # Label-oriented
Out[49]:
AAA BBB CCC
1 4 10 100
2 5 20 50
3 6 30 -30In [50]: df = pd.DataFrame(
....: {"AAA": [4, 5, 6, 7], "BBB": [10, 20, 30, 40], "CCC": [100, 50, -30, -50]}
....: )
....:
In [51]: df
Out[51]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [52]: df[~((df.AAA <= 6) & (df.index.isin([0, 2, 4])))]
Out[52]:
AAA BBB CCC
1 5 20 50
3 7 40 -50New columns
In [53]: df = pd.DataFrame({"AAA": [1, 2, 1, 3], "BBB": [1, 1, 2, 2], "CCC": [2, 1, 3, 1]})
In [54]: df
Out[54]:
AAA BBB CCC
0 1 1 2
1 2 1 1
2 1 2 3
3 3 2 1
In [55]: source_cols = df.columns # Or some subset would work too
In [56]: new_cols = [str(x) + "_cat" for x in source_cols]
In [57]: categories = {1: "Alpha", 2: "Beta", 3: "Charlie"}
In [58]: df[new_cols] = df[source_cols].map(categories.get)
In [59]: df
Out[59]:
AAA BBB CCC AAA_cat BBB_cat CCC_cat
0 1 1 2 Alpha Alpha Beta
1 2 1 1 Beta Alpha Alpha
2 1 2 3 Alpha Beta Charlie
3 3 2 1 Charlie Beta AlphaIn [60]: df = pd.DataFrame(
....: {"AAA": [1, 1, 1, 2, 2, 2, 3, 3], "BBB": [2, 1, 3, 4, 5, 1, 2, 3]}
....: )
....:
In [61]: df
Out[61]:
AAA BBB
0 1 2
1 1 1
2 1 3
3 2 4
4 2 5
5 2 1
6 3 2
7 3 3方法 1：使用 idxmin() 得到最小值的索引
Method 1 : idxmin() to get the index of the minimums
In [62]: df.loc[df.groupby("AAA")["BBB"].idxmin()]
Out[62]:
AAA BBB
1 1 1
5 2 1
6 3 2方法 2：排序然后获取每个的第一个
Method 2 : sort then take first of each
In [63]: df.sort_values(by="BBB").groupby("AAA", as_index=False).first()
Out[63]:
AAA BBB
0 1 1
1 2 1
2 3 2注意，除了索引之外，结果相同。
Notice the same results, with the exception of the index.
Multiindexing
multindexing 文档。
The multindexing docs.
In [64]: df = pd.DataFrame(
....: {
....: "row": [0, 1, 2],
....: "One_X": [1.1, 1.1, 1.1],
....: "One_Y": [1.2, 1.2, 1.2],
....: "Two_X": [1.11, 1.11, 1.11],
....: "Two_Y": [1.22, 1.22, 1.22],
....: }
....: )
....:
In [65]: df
Out[65]:
row One_X One_Y Two_X Two_Y
0 0 1.1 1.2 1.11 1.22
1 1 1.1 1.2 1.11 1.22
2 2 1.1 1.2 1.11 1.22
# As Labelled Index
In [66]: df = df.set_index("row")
In [67]: df
Out[67]:
One_X One_Y Two_X Two_Y
row
0 1.1 1.2 1.11 1.22
1 1.1 1.2 1.11 1.22
2 1.1 1.2 1.11 1.22
# With Hierarchical Columns
In [68]: df.columns = pd.MultiIndex.from_tuples([tuple(c.split("_")) for c in df.columns])
In [69]: df
Out[69]:
One Two
X Y X Y
row
0 1.1 1.2 1.11 1.22
1 1.1 1.2 1.11 1.22
2 1.1 1.2 1.11 1.22
# Now stack & Reset
In [70]: df = df.stack(0, future_stack=True).reset_index(1)
In [71]: df
Out[71]:
level_1 X Y
row
0 One 1.10 1.20
0 Two 1.11 1.22
1 One 1.10 1.20
1 Two 1.11 1.22
2 One 1.10 1.20
2 Two 1.11 1.22
# And fix the labels (Notice the label 'level_1' got added automatically)
In [72]: df.columns = ["Sample", "All_X", "All_Y"]
In [73]: df
Out[73]:
Sample All_X All_Y
row
0 One 1.10 1.20
0 Two 1.11 1.22
1 One 1.10 1.20
1 Two 1.11 1.22
2 One 1.10 1.20
2 Two 1.11 1.22Arithmetic
In [74]: cols = pd.MultiIndex.from_tuples(
....: [(x, y) for x in ["A", "B", "C"] for y in ["O", "I"]]
....: )
....:
In [75]: df = pd.DataFrame(np.random.randn(2, 6), index=["n", "m"], columns=cols)
In [76]: df
Out[76]:
A B C
O I O I O I
n 0.469112 -0.282863 -1.509059 -1.135632 1.212112 -0.173215
m 0.119209 -1.044236 -0.861849 -2.104569 -0.494929 1.071804
In [77]: df = df.div(df["C"], level=1)
In [78]: df
Out[78]:
A B C
O I O I O I
n 0.387021 1.633022 -1.244983 6.556214 1.0 1.0
m -0.240860 -0.974279 1.741358 -1.963577 1.0 1.0Slicing
In [79]: coords = [("AA", "one"), ("AA", "six"), ("BB", "one"), ("BB", "two"), ("BB", "six")]
In [80]: index = pd.MultiIndex.from_tuples(coords)
In [81]: df = pd.DataFrame([11, 22, 33, 44, 55], index, ["MyData"])
In [82]: df
Out[82]:
MyData
AA one 11
six 22
BB one 33
two 44
six 55要对第一层的第一轴的索引进行交叉部分：
To take the cross section of the 1st level and 1st axis the index:
# Note : level and axis are optional, and default to zero
In [83]: df.xs("BB", level=0, axis=0)
Out[83]:
MyData
one 33
two 44
six 55… 现在是第一轴的第二层。
…and now the 2nd level of the 1st axis.
In [84]: df.xs("six", level=1, axis=0)
Out[84]:
MyData
AA 22
BB 55In [85]: import itertools
In [86]: index = list(itertools.product(["Ada", "Quinn", "Violet"], ["Comp", "Math", "Sci"]))
In [87]: headr = list(itertools.product(["Exams", "Labs"], ["I", "II"]))
In [88]: indx = pd.MultiIndex.from_tuples(index, names=["Student", "Course"])
In [89]: cols = pd.MultiIndex.from_tuples(headr) # Notice these are un-named
In [90]: data = [[70 + x + y + (x * y) % 3 for x in range(4)] for y in range(9)]
In [91]: df = pd.DataFrame(data, indx, cols)
In [92]: df
Out[92]:
Exams Labs
I II I II
Student Course
Ada Comp 70 71 72 73
Math 71 73 75 74
Sci 72 75 75 75
Quinn Comp 73 74 75 76
Math 74 76 78 77
Sci 75 78 78 78
Violet Comp 76 77 78 79
Math 77 79 81 80
Sci 78 81 81 81
In [93]: All = slice(None)
In [94]: df.loc["Violet"]
Out[94]:
Exams Labs
I II I II
Course
Comp 76 77 78 79
Math 77 79 81 80
Sci 78 81 81 81
In [95]: df.loc[(All, "Math"), All]
Out[95]:
Exams Labs
I II I II
Student Course
Ada Math 71 73 75 74
Quinn Math 74 76 78 77
Violet Math 77 79 81 80
In [96]: df.loc[(slice("Ada", "Quinn"), "Math"), All]
Out[96]:
Exams Labs
I II I II
Student Course
Ada Math 71 73 75 74
Quinn Math 74 76 78 77
In [97]: df.loc[(All, "Math"), ("Exams")]
Out[97]:
I II
Student Course
Ada Math 71 73
Quinn Math 74 76
Violet Math 77 79
In [98]: df.loc[(All, "Math"), (All, "II")]
Out[98]:
Exams Labs
II II
Student Course
Ada Math 73 74
Quinn Math 76 77
Violet Math 79 80Sorting
In [99]: df.sort_values(by=("Labs", "II"), ascending=False)
Out[99]:
Exams Labs
I II I II
Student Course
Violet Sci 78 81 81 81
Math 77 79 81 80
Comp 76 77 78 79
Quinn Sci 75 78 78 78
Math 74 76 78 77
Comp 73 74 75 76
Ada Sci 72 75 75 75
Math 71 73 75 74
Comp 70 71 72 73部分选择，有序性的需要 GH 2995
Partial selection, the need for sortedness GH 2995
Missing data
missing data 文档。
The missing data docs.
向前填充一个反向时序
Fill forward a reversed timeseries
In [100]: df = pd.DataFrame(
.....: np.random.randn(6, 1),
.....: index=pd.date_range("2013-08-01", periods=6, freq="B"),
.....: columns=list("A"),
.....: )
.....:
In [101]: df.loc[df.index[3], "A"] = np.nan
In [102]: df
Out[102]:
A
2013-08-01 0.721555
2013-08-02 -0.706771
2013-08-05 -1.039575
2013-08-06 NaN
2013-08-07 -0.424972
2013-08-08 0.567020
In [103]: df.bfill()
Out[103]:
A
2013-08-01 0.721555
2013-08-02 -0.706771
2013-08-05 -1.039575
2013-08-06 -0.424972
2013-08-07 -0.424972
2013-08-08 0.567020Grouping
grouping 文档。
The grouping docs.
与 agg 不同，apply 的可调用对象会传递一个子 DataFrame，让你可以访问所有列
Unlike agg, apply’s callable is passed a sub-DataFrame which gives you access to all the columns
In [104]: df = pd.DataFrame(
.....: {
.....: "animal": "cat dog cat fish dog cat cat".split(),
.....: "size": list("SSMMMLL"),
.....: "weight": [8, 10, 11, 1, 20, 12, 12],
.....: "adult": [False] * 5 + [True] * 2,
.....: }
.....: )
.....:
In [105]: df
Out[105]:
animal size weight adult
0 cat S 8 False
1 dog S 10 False
2 cat M 11 False
3 fish M 1 False
4 dog M 20 False
5 cat L 12 True
6 cat L 12 True
# List the size of the animals with the highest weight.
In [106]: df.groupby("animal").apply(lambda subf: subf["size"][subf["weight"].idxmax()], include_groups=False)
Out[106]:
animal
cat L
dog M
fish M
dtype: objectIn [107]: gb = df.groupby("animal")
In [108]: gb.get_group("cat")
Out[108]:
animal size weight adult
0 cat S 8 False
2 cat M 11 False
5 cat L 12 True
6 cat L 12 TrueIn [109]: def GrowUp(x):
.....: avg_weight = sum(x[x["size"] == "S"].weight * 1.5)
.....: avg_weight += sum(x[x["size"] == "M"].weight * 1.25)
.....: avg_weight += sum(x[x["size"] == "L"].weight)
.....: avg_weight /= len(x)
.....: return pd.Series(["L", avg_weight, True], index=["size", "weight", "adult"])
.....:
In [110]: expected_df = gb.apply(GrowUp, include_groups=False)
In [111]: expected_df
Out[111]:
size weight adult
animal
cat L 12.4375 True
dog L 20.0000 True
fish L 1.2500 TrueIn [112]: S = pd.Series([i / 100.0 for i in range(1, 11)])
In [113]: def cum_ret(x, y):
.....: return x * (1 + y)
.....:
In [114]: def red(x):
.....: return functools.reduce(cum_ret, x, 1.0)
.....:
In [115]: S.expanding().apply(red, raw=True)
Out[115]:
0 1.010000
1 1.030200
2 1.061106
3 1.103550
4 1.158728
5 1.228251
6 1.314229
7 1.419367
8 1.547110
9 1.701821
dtype: float64In [116]: df = pd.DataFrame({"A": [1, 1, 2, 2], "B": [1, -1, 1, 2]})
In [117]: gb = df.groupby("A")
In [118]: def replace(g):
.....: mask = g < 0
.....: return g.where(~mask, g[~mask].mean())
.....:
In [119]: gb.transform(replace)
Out[119]:
B
0 1
1 1
2 1
3 2In [120]: df = pd.DataFrame(
.....: {
.....: "code": ["foo", "bar", "baz"] * 2,
.....: "data": [0.16, -0.21, 0.33, 0.45, -0.59, 0.62],
.....: "flag": [False, True] * 3,
.....: }
.....: )
.....:
In [121]: code_groups = df.groupby("code")
In [122]: agg_n_sort_order = code_groups[["data"]].transform("sum").sort_values(by="data")
In [123]: sorted_df = df.loc[agg_n_sort_order.index]
In [124]: sorted_df
Out[124]:
code data flag
1 bar -0.21 True
4 bar -0.59 False
0 foo 0.16 False
3 foo 0.45 True
2 baz 0.33 False
5 baz 0.62 TrueIn [125]: rng = pd.date_range(start="2014-10-07", periods=10, freq="2min")
In [126]: ts = pd.Series(data=list(range(10)), index=rng)
In [127]: def MyCust(x):
.....: if len(x) > 2:
.....: return x.iloc[1] * 1.234
.....: return pd.NaT
.....:
In [128]: mhc = {"Mean": "mean", "Max": "max", "Custom": MyCust}
In [129]: ts.resample("5min").apply(mhc)
Out[129]:
Mean Max Custom
2014-10-07 00:00:00 1.0 2 1.234
2014-10-07 00:05:00 3.5 4 NaT
2014-10-07 00:10:00 6.0 7 7.404
2014-10-07 00:15:00 8.5 9 NaT
In [130]: ts
Out[130]:
2014-10-07 00:00:00 0
2014-10-07 00:02:00 1
2014-10-07 00:04:00 2
2014-10-07 00:06:00 3
2014-10-07 00:08:00 4
2014-10-07 00:10:00 5
2014-10-07 00:12:00 6
2014-10-07 00:14:00 7
2014-10-07 00:16:00 8
2014-10-07 00:18:00 9
Freq: 2min, dtype: int64In [131]: df = pd.DataFrame(
.....: {"Color": "Red Red Red Blue".split(), "Value": [100, 150, 50, 50]}
.....: )
.....:
In [132]: df
Out[132]:
Color Value
0 Red 100
1 Red 150
2 Red 50
3 Blue 50
In [133]: df["Counts"] = df.groupby(["Color"]).transform(len)
In [134]: df
Out[134]:
Color Value Counts
0 Red 100 3
1 Red 150 3
2 Red 50 3
3 Blue 50 1In [135]: df = pd.DataFrame(
.....: {"line_race": [10, 10, 8, 10, 10, 8], "beyer": [99, 102, 103, 103, 88, 100]},
.....: index=[
.....: "Last Gunfighter",
.....: "Last Gunfighter",
.....: "Last Gunfighter",
.....: "Paynter",
.....: "Paynter",
.....: "Paynter",
.....: ],
.....: )
.....:
In [136]: df
Out[136]:
line_race beyer
Last Gunfighter 10 99
Last Gunfighter 10 102
Last Gunfighter 8 103
Paynter 10 103
Paynter 10 88
Paynter 8 100
In [137]: df["beyer_shifted"] = df.groupby(level=0)["beyer"].shift(1)
In [138]: df
Out[138]:
line_race beyer beyer_shifted
Last Gunfighter 10 99 NaN
Last Gunfighter 10 102 99.0
Last Gunfighter 8 103 102.0
Paynter 10 103 NaN
Paynter 10 88 103.0
Paynter 8 100 88.0In [139]: df = pd.DataFrame(
.....: {
.....: "host": ["other", "other", "that", "this", "this"],
.....: "service": ["mail", "web", "mail", "mail", "web"],
.....: "no": [1, 2, 1, 2, 1],
.....: }
.....: ).set_index(["host", "service"])
.....:
In [140]: mask = df.groupby(level=0).agg("idxmax")
In [141]: df_count = df.loc[mask["no"]].reset_index()
In [142]: df_count
Out[142]:
host service no
0 other web 2
1 that mail 1
2 this mail 2In [143]: df = pd.DataFrame([0, 1, 0, 1, 1, 1, 0, 1, 1], columns=["A"])
In [144]: df["A"].groupby((df["A"] != df["A"].shift()).cumsum()).groups
Out[144]: {1: [0], 2: [1], 3: [2], 4: [3, 4, 5], 5: [6], 6: [7, 8]}
In [145]: df["A"].groupby((df["A"] != df["A"].shift()).cumsum()).cumsum()
Out[145]:
0 0
1 1
2 0
3 1
4 2
5 3
6 0
7 1
8 2
Name: A, dtype: int64Splitting
创建一个数据框列表，使用基于包含在行中的逻辑的划分进行拆分。
Create a list of dataframes, split using a delineation based on logic included in rows.
In [146]: df = pd.DataFrame(
.....: data={
.....: "Case": ["A", "A", "A", "B", "A", "A", "B", "A", "A"],
.....: "Data": np.random.randn(9),
.....: }
.....: )
.....:
In [147]: dfs = list(
.....: zip(
.....: *df.groupby(
.....: (1 * (df["Case"] == "B"))
.....: .cumsum()
.....: .rolling(window=3, min_periods=1)
.....: .median()
.....: )
.....: )
.....: )[-1]
.....:
In [148]: dfs[0]
Out[148]:
Case Data
0 A 0.276232
1 A -1.087401
2 A -0.673690
3 B 0.113648
In [149]: dfs[1]
Out[149]:
Case Data
4 A -1.478427
5 A 0.524988
6 B 0.404705
In [150]: dfs[2]
Out[150]:
Case Data
7 A 0.577046
8 A -1.715002Pivot
Pivot 文档。
The Pivot docs.
In [151]: df = pd.DataFrame(
.....: data={
.....: "Province": ["ON", "QC", "BC", "AL", "AL", "MN", "ON"],
.....: "City": [
.....: "Toronto",
.....: "Montreal",
.....: "Vancouver",
.....: "Calgary",
.....: "Edmonton",
.....: "Winnipeg",
.....: "Windsor",
.....: ],
.....: "Sales": [13, 6, 16, 8, 4, 3, 1],
.....: }
.....: )
.....:
In [152]: table = pd.pivot_table(
.....: df,
.....: values=["Sales"],
.....: index=["Province"],
.....: columns=["City"],
.....: aggfunc="sum",
.....: margins=True,
.....: )
.....:
In [153]: table.stack("City", future_stack=True)
Out[153]:
Sales
Province City
AL Calgary 8.0
Edmonton 4.0
Montreal NaN
Toronto NaN
Vancouver NaN
... ...
All Toronto 13.0
Vancouver 16.0
Windsor 1.0
Winnipeg 3.0
All 51.0
[48 rows x 1 columns]In [154]: grades = [48, 99, 75, 80, 42, 80, 72, 68, 36, 78]
In [155]: df = pd.DataFrame(
.....: {
.....: "ID": ["x%d" % r for r in range(10)],
.....: "Gender": ["F", "M", "F", "M", "F", "M", "F", "M", "M", "M"],
.....: "ExamYear": [
.....: "2007",
.....: "2007",
.....: "2007",
.....: "2008",
.....: "2008",
.....: "2008",
.....: "2008",
.....: "2009",
.....: "2009",
.....: "2009",
.....: ],
.....: "Class": [
.....: "algebra",
.....: "stats",
.....: "bio",
.....: "algebra",
.....: "algebra",
.....: "stats",
.....: "stats",
.....: "algebra",
.....: "bio",
.....: "bio",
.....: ],
.....: "Participated": [
.....: "yes",
.....: "yes",
.....: "yes",
.....: "yes",
.....: "no",
.....: "yes",
.....: "yes",
.....: "yes",
.....: "yes",
.....: "yes",
.....: ],
.....: "Passed": ["yes" if x > 50 else "no" for x in grades],
.....: "Employed": [
.....: True,
.....: True,
.....: True,
.....: False,
.....: False,
.....: False,
.....: False,
.....: True,
.....: True,
.....: False,
.....: ],
.....: "Grade": grades,
.....: }
.....: )
.....:
In [156]: df.groupby("ExamYear").agg(
.....: {
.....: "Participated": lambda x: x.value_counts()["yes"],
.....: "Passed": lambda x: sum(x == "yes"),
.....: "Employed": lambda x: sum(x),
.....: "Grade": lambda x: sum(x) / len(x),
.....: }
.....: )
.....:
Out[156]:
Participated Passed Employed Grade
ExamYear
2007 3 2 3 74.000000
2008 3 3 0 68.500000
2009 3 2 2 60.666667要创建年度和月份交叉制表：
To create year and month cross tabulation:
In [157]: df = pd.DataFrame(
.....: {"value": np.random.randn(36)},
.....: index=pd.date_range("2011-01-01", freq="ME", periods=36),
.....: )
.....:
In [158]: pd.pivot_table(
.....: df, index=df.index.month, columns=df.index.year, values="value", aggfunc="sum"
.....: )
.....:
Out[158]:
2011 2012 2013
1 -1.039268 -0.968914 2.565646
2 -0.370647 -1.294524 1.431256
3 -1.157892 0.413738 1.340309
4 -1.344312 0.276662 -1.170299
5 0.844885 -0.472035 -0.226169
6 1.075770 -0.013960 0.410835
7 -0.109050 -0.362543 0.813850
8 1.643563 -0.006154 0.132003
9 -1.469388 -0.923061 -0.827317
10 0.357021 0.895717 -0.076467
11 -0.674600 0.805244 -1.187678
12 -1.776904 -1.206412 1.130127Apply
In [159]: df = pd.DataFrame(
.....: data={
.....: "A": [[2, 4, 8, 16], [100, 200], [10, 20, 30]],
.....: "B": [["a", "b", "c"], ["jj", "kk"], ["ccc"]],
.....: },
.....: index=["I", "II", "III"],
.....: )
.....:
In [160]: def SeriesFromSubList(aList):
.....: return pd.Series(aList)
.....:
In [161]: df_orgz = pd.concat(
.....: {ind: row.apply(SeriesFromSubList) for ind, row in df.iterrows()}
.....: )
.....:
In [162]: df_orgz
Out[162]:
0 1 2 3
I A 2 4 8 16.0
B a b c NaN
II A 100 200 NaN NaN
B jj kk NaN NaN
III A 10 20.0 30.0 NaN
B ccc NaN NaN NaN滚动应用到多列，其中函数在从 Series 返回标量之前计算 Series
Rolling Apply to multiple columns where function calculates a Series before a Scalar from the Series is returned
In [163]: df = pd.DataFrame(
.....: data=np.random.randn(2000, 2) / 10000,
.....: index=pd.date_range("2001-01-01", periods=2000),
.....: columns=["A", "B"],
.....: )
.....:
In [164]: df
Out[164]:
A B
2001-01-01 -0.000144 -0.000141
2001-01-02 0.000161 0.000102
2001-01-03 0.000057 0.000088
2001-01-04 -0.000221 0.000097
2001-01-05 -0.000201 -0.000041
... ... ...
2006-06-19 0.000040 -0.000235
2006-06-20 -0.000123 -0.000021
2006-06-21 -0.000113 0.000114
2006-06-22 0.000136 0.000109
2006-06-23 0.000027 0.000030
[2000 rows x 2 columns]
In [165]: def gm(df, const):
.....: v = ((((df["A"] + df["B"]) + 1).cumprod()) - 1) * const
.....: return v.iloc[-1]
.....:
In [166]: s = pd.Series(
.....: {
.....: df.index[i]: gm(df.iloc[i: min(i + 51, len(df) - 1)], 5)
.....: for i in range(len(df) - 50)
.....: }
.....: )
.....:
In [167]: s
Out[167]:
2001-01-01 0.000930
2001-01-02 0.002615
2001-01-03 0.001281
2001-01-04 0.001117
2001-01-05 0.002772
...
2006-04-30 0.003296
2006-05-01 0.002629
2006-05-02 0.002081
2006-05-03 0.004247
2006-05-04 0.003928
Length: 1950, dtype: float64滚动应用到多列，其中函数返回标量（成交量加权平均价）
Rolling Apply to multiple columns where function returns a Scalar (Volume Weighted Average Price)
In [168]: rng = pd.date_range(start="2014-01-01", periods=100)
In [169]: df = pd.DataFrame(
.....: {
.....: "Open": np.random.randn(len(rng)),
.....: "Close": np.random.randn(len(rng)),
.....: "Volume": np.random.randint(100, 2000, len(rng)),
.....: },
.....: index=rng,
.....: )
.....:
In [170]: df
Out[170]:
Open Close Volume
2014-01-01 -1.611353 -0.492885 1219
2014-01-02 -3.000951 0.445794 1054
2014-01-03 -0.138359 -0.076081 1381
2014-01-04 0.301568 1.198259 1253
2014-01-05 0.276381 -0.669831 1728
... ... ... ...
2014-04-06 -0.040338 0.937843 1188
2014-04-07 0.359661 -0.285908 1864
2014-04-08 0.060978 1.714814 941
2014-04-09 1.759055 -0.455942 1065
2014-04-10 0.138185 -1.147008 1453
[100 rows x 3 columns]
In [171]: def vwap(bars):
.....: return (bars.Close * bars.Volume).sum() / bars.Volume.sum()
.....:
In [172]: window = 5
In [173]: s = pd.concat(
.....: [
.....: (pd.Series(vwap(df.iloc[i: i + window]), index=[df.index[i + window]]))
.....: for i in range(len(df) - window)
.....: ]
.....: )
.....:
In [174]: s.round(2)
Out[174]:
2014-01-06 0.02
2014-01-07 0.11
2014-01-08 0.10
2014-01-09 0.07
2014-01-10 -0.29
...
2014-04-06 -0.63
2014-04-07 -0.02
2014-04-08 -0.03
2014-04-09 0.34
2014-04-10 0.29
Length: 95, dtype: float64Timeseries
将以列为小时数和以行为天数的矩阵，变成一个以时间序列的形式连续的行序列。 How to rearrange a Python pandas DataFrame?
Turn a matrix with hours in columns and days in rows into a continuous row sequence in the form of a time series. How to rearrange a Python pandas DataFrame?
计算 DatetimeIndex 中每个条目所在月份的第一天
Calculate the first day of the month for each entry in a DatetimeIndex
In [175]: dates = pd.date_range("2000-01-01", periods=5)
In [176]: dates.to_period(freq="M").to_timestamp()
Out[176]:
DatetimeIndex(['2000-01-01', '2000-01-01', '2000-01-01', '2000-01-01',
'2000-01-01'],
dtype='datetime64[ns]', freq=None)Resampling
Resample 文档。
The Resample docs.
Grouper 中有效的频率参数 Timeseries
Valid frequency arguments to Grouper Timeseries
使用 TimeGrouper 和另一个分组创建子组，然后应用自定义函数 GH 3791
Using TimeGrouper and another grouping to create subgroups, then apply a custom function GH 3791
Merge
Join 文档。
The Join docs.
In [177]: rng = pd.date_range("2000-01-01", periods=6)
In [178]: df1 = pd.DataFrame(np.random.randn(6, 3), index=rng, columns=["A", "B", "C"])
In [179]: df2 = df1.copy()根据 df 构造，ignore_index 可能必须使用
Depending on df construction, ignore_index may be needed
In [180]: df = pd.concat([df1, df2], ignore_index=True)
In [181]: df
Out[181]:
A B C
0 -0.870117 -0.479265 -0.790855
1 0.144817 1.726395 -0.464535
2 -0.821906 1.597605 0.187307
3 -0.128342 -1.511638 -0.289858
4 0.399194 -1.430030 -0.639760
5 1.115116 -2.012600 1.810662
6 -0.870117 -0.479265 -0.790855
7 0.144817 1.726395 -0.464535
8 -0.821906 1.597605 0.187307
9 -0.128342 -1.511638 -0.289858
10 0.399194 -1.430030 -0.639760
11 1.115116 -2.012600 1.810662DataFrame 的自连接 GH 2996
Self Join of a DataFrame GH 2996
In [182]: df = pd.DataFrame(
.....: data={
.....: "Area": ["A"] * 5 + ["C"] * 2,
.....: "Bins": [110] * 2 + [160] * 3 + [40] * 2,
.....: "Test_0": [0, 1, 0, 1, 2, 0, 1],
.....: "Data": np.random.randn(7),
.....: }
.....: )
.....:
In [183]: df
Out[183]:
Area Bins Test_0 Data
0 A 110 0 -0.433937
1 A 110 1 -0.160552
2 A 160 0 0.744434
3 A 160 1 1.754213
4 A 160 2 0.000850
5 C 40 0 0.342243
6 C 40 1 1.070599
In [184]: df["Test_1"] = df["Test_0"] - 1
In [185]: pd.merge(
.....: df,
.....: df,
.....: left_on=["Bins", "Area", "Test_0"],
.....: right_on=["Bins", "Area", "Test_1"],
.....: suffixes=("_L", "_R"),
.....: )
.....:
Out[185]:
Area Bins Test_0_L Data_L Test_1_L Test_0_R Data_R Test_1_R
0 A 110 0 -0.433937 -1 1 -0.160552 0
1 A 160 0 0.744434 -1 1 1.754213 0
2 A 160 1 1.754213 0 2 0.000850 1
3 C 40 0 0.342243 -1 1 1.070599 0Plotting
Plotting 文档。
The Plotting docs.
In [186]: df = pd.DataFrame(
.....: {
.....: "stratifying_var": np.random.uniform(0, 100, 20),
.....: "price": np.random.normal(100, 5, 20),
.....: }
.....: )
.....:
In [187]: df["quartiles"] = pd.qcut(
.....: df["stratifying_var"], 4, labels=["0-25%", "25-50%", "50-75%", "75-100%"]
.....: )
.....:
In [188]: df.boxplot(column="price", by="quartiles")
Out[188]: <Axes: title={'center': 'price'}, xlabel='quartiles'>Data in/out
CSV
CSV 文档
The CSV docs
读取一个已压缩但非由 gzip/bz2 压缩的文件（其为 read_csv 识别的压缩格式）。此示例显示了一个 WinZipped 文件，但其可作为使用上下文管理器打开文件并在其中使用该句柄进行读取的通用应用程序。 See here
Reading a file that is compressed but not by gzip/bz2 (the native compressed formats which read_csv understands). This example shows a WinZipped file, but is a general application of opening the file within a context manager and using that handle to read. See here
处理不当的线条 GH 2886
Dealing with bad lines GH 2886
将多个文件合并到单个 DataFrame 中的最好方法是逐个读取各个帧，将所有单个帧放入列表，然后使用 pd.concat() 合并列表中的帧：
The best way to combine multiple files into a single DataFrame is to read the individual frames one by one, put all of the individual frames into a list, and then combine the frames in the list using pd.concat():
In [189]: for i in range(3):
.....: data = pd.DataFrame(np.random.randn(10, 4))
.....: data.to_csv("file_{}.csv".format(i))
.....:
In [190]: files = ["file_0.csv", "file_1.csv", "file_2.csv"]
In [191]: result = pd.concat([pd.read_csv(f) for f in files], ignore_index=True)你可以使用相同的方法读取匹配模式的所有文件。下面是一个使用 glob 的示例：
You can use the same approach to read all files matching a pattern. Here is an example using glob:
In [192]: import glob
In [193]: import os
In [194]: files = glob.glob("file_*.csv")
In [195]: result = pd.concat([pd.read_csv(f) for f in files], ignore_index=True)最后，此策略将与 io docs 中描述的其他 pd.read*(…)_ 函数配合使用。
Finally, this strategy will work with the other pd.read*(…)_ functions described in the io docs.
在多列中解析日期组件使用格式更快
Parsing date components in multi-columns is faster with a format
In [196]: i = pd.date_range("20000101", periods=10000)
In [197]: df = pd.DataFrame({"year": i.year, "month": i.month, "day": i.day})
In [198]: df.head()
Out[198]:
year month day
0 2000 1 1
1 2000 1 2
2 2000 1 3
3 2000 1 4
4 2000 1 5
In [199]: %timeit pd.to_datetime(df.year * 10000 + df.month * 100 + df.day, format='%Y%m%d')
.....: ds = df.apply(lambda x: "%04d%02d%02d" % (x["year"], x["month"], x["day"]), axis=1)
.....: ds.head()
.....: %timeit pd.to_datetime(ds)
.....:
4.01 ms +- 635 us per loop (mean +- std. dev. of 7 runs, 100 loops each)
1.05 ms +- 7.39 us per loop (mean +- std. dev. of 7 runs, 1,000 loops each)In [200]: data = """;;;;
.....: ;;;;
.....: ;;;;
.....: ;;;;
.....: ;;;;
.....: ;;;;
.....: ;;;;
.....: ;;;;
.....: ;;;;
.....: ;;;;
.....: date;Param1;Param2;Param4;Param5
.....: ;m²;°C;m²;m
.....: ;;;;
.....: 01.01.1990 00:00;1;1;2;3
.....: 01.01.1990 01:00;5;3;4;5
.....: 01.01.1990 02:00;9;5;6;7
.....: 01.01.1990 03:00;13;7;8;9
.....: 01.01.1990 04:00;17;9;10;11
.....: 01.01.1990 05:00;21;11;12;13
.....: """
.....:In [201]: from io import StringIO
In [202]: pd.read_csv(
.....: StringIO(data),
.....: sep=";",
.....: skiprows=[11, 12],
.....: index_col=0,
.....: parse_dates=True,
.....: header=10,
.....: )
.....:
Out[202]:
Param1 Param2 Param4 Param5
date
1990-01-01 00:00:00 1 1 2 3
1990-01-01 01:00:00 5 3 4 5
1990-01-01 02:00:00 9 5 6 7
1990-01-01 03:00:00 13 7 8 9
1990-01-01 04:00:00 17 9 10 11
1990-01-01 05:00:00 21 11 12 13In [203]: pd.read_csv(StringIO(data), sep=";", header=10, nrows=10).columns
Out[203]: Index(['date', 'Param1', 'Param2', 'Param4', 'Param5'], dtype='object')
In [204]: columns = pd.read_csv(StringIO(data), sep=";", header=10, nrows=10).columns
In [205]: pd.read_csv(
.....: StringIO(data), sep=";", index_col=0, header=12, parse_dates=True, names=columns
.....: )
.....:
Out[205]:
Param1 Param2 Param4 Param5
date
1990-01-01 00:00:00 1 1 2 3
1990-01-01 01:00:00 5 3 4 5
1990-01-01 02:00:00 9 5 6 7
1990-01-01 03:00:00 13 7 8 9
1990-01-01 04:00:00 17 9 10 11
1990-01-01 05:00:00 21 11 12 13Excel
Excel 文档
The Excel docs
仅加载可见的表单 GH 19842#issuecomment-892150745
Loading only visible sheets GH 19842#issuecomment-892150745
HDFStore
HDFStores 文档
The HDFStores docs
使用链接的多表层次结构管理异构数据 GH 3032
Managing heterogeneous data using a linked multiple table hierarchy GH 3032
使用 Gemini 将其从大份数据块中去重，基本上是一个递归的归约操作。展示了一个从 csv 文件中获取数据并通过数据块创建一个存储功能，还可以解析日期。 See here
De-duplicating a large store by chunks, essentially a recursive reduction operation. Shows a function for taking in data from csv file and creating a store by chunks, with date parsing as well. See here
将属性存储到组节点
Storing Attributes to a group node
In [206]: df = pd.DataFrame(np.random.randn(8, 3))
In [207]: store = pd.HDFStore("test.h5")
In [208]: store.put("df", df)
# you can store an arbitrary Python object via pickle
In [209]: store.get_storer("df").attrs.my_attribute = {"A": 10}
In [210]: store.get_storer("df").attrs.my_attribute
Out[210]: {'A': 10}您可以通过将 driver 参数传递给 PyTables 来在内存中创建或加载 HDFStore。只有当 HDFStore 关闭时才会将更改写入磁盘。
You can create or load a HDFStore in-memory by passing the driver parameter to PyTables. Changes are only written to disk when the HDFStore is closed.
In [211]: store = pd.HDFStore("test.h5", "w", driver="H5FD_CORE")
In [212]: df = pd.DataFrame(np.random.randn(8, 3))
In [213]: store["test"] = df
# only after closing the store, data is written to disk:
In [214]: store.close()Binary files
如果您需要读取由 C 结构数组组成的二进制文件，则 pandas 就能轻松接受 NumPy 记录数组。例如，给定一个名为 main.c 的文件中用 gcc main.c -std=gnu99 在 64 位计算机上编译的 C 程序，
pandas readily accepts NumPy record arrays, if you need to read in a binary file consisting of an array of C structs. For example, given this C program in a file called main.c compiled with gcc main.c -std=gnu99 on a 64-bit machine,
#include <stdio.h>
#include <stdint.h>
typedef struct _Data
{
int32_t count;
double avg;
float scale;
} Data;
int main(int argc, const char *argv[])
{
size_t n = 10;
Data d[n];
for (int i = 0; i < n; ++i)
{
d[i].count = i;
d[i].avg = i + 1.0;
d[i].scale = (float) i + 2.0f;
}
FILE *file = fopen("binary.dat", "wb");
fwrite(&d, sizeof(Data), n, file);
fclose(file);
return 0;
}如下的 Python 代码会将二进制文件 'binary.dat' 读取到 pandas DataFrame 中，其中结构的每个元素都对应于框架中的一个列：
the following Python code will read the binary file 'binary.dat' into a pandas DataFrame, where each element of the struct corresponds to a column in the frame:
names = "count", "avg", "scale"
# note that the offsets are larger than the size of the type because of
# struct padding
offsets = 0, 8, 16
formats = "i4", "f8", "f4"
dt = np.dtype({"names": names, "offsets": offsets, "formats": formats}, align=True)
df = pd.DataFrame(np.fromfile("binary.dat", dt))|
结构元素的偏移量可能会根据创建文件时所用机器的架构而有所不同。不建议将这样的原始二进制文件格式作为一般数据存储使用，因为它不是跨平台的。我们建议使用 HDF5 或 parquet，它们都受 pandas 的 IO 工具支持。 |
|
The offsets of the structure elements may be different depending on the architecture of the machine on which the file was created. Using a raw binary file format like this for general data storage is not recommended, as it is not cross platform. We recommended either HDF5 or parquet, both of which are supported by pandas’ IO facilities. |
Computation
Correlation
通常情况下，从 DataFrame.corr() 计算出的相关系数矩阵的以下 (或上) 三角形形式是有用的。通过将一个布尔掩码传递给 where，可以实现这一点：
Often it’s useful to obtain the lower (or upper) triangular form of a correlation matrix calculated from DataFrame.corr(). This can be achieved by passing a boolean mask to where as follows:
In [215]: df = pd.DataFrame(np.random.random(size=(100, 5)))
In [216]: corr_mat = df.corr()
In [217]: mask = np.tril(np.ones_like(corr_mat, dtype=np.bool_), k=-1)
In [218]: corr_mat.where(mask)
Out[218]:
0 1 2 3 4
0 NaN NaN NaN NaN NaN
1 -0.079861 NaN NaN NaN NaN
2 -0.236573 0.183801 NaN NaN NaN
3 -0.013795 -0.051975 0.037235 NaN NaN
4 -0.031974 0.118342 -0.073499 -0.02063 NaNmethod 中的 DataFrame.corr 参数接受可调用的内容，此外还可以接受命名的关联类型。在这里，我们计算 DataFrame 对象的 distance correlation 矩阵。
The method argument within DataFrame.corr can accept a callable in addition to the named correlation types. Here we compute the distance correlation matrix for a DataFrame object.
In [219]: def distcorr(x, y):
.....: n = len(x)
.....: a = np.zeros(shape=(n, n))
.....: b = np.zeros(shape=(n, n))
.....: for i in range(n):
.....: for j in range(i + 1, n):
.....: a[i, j] = abs(x[i] - x[j])
.....: b[i, j] = abs(y[i] - y[j])
.....: a += a.T
.....: b += b.T
.....: a_bar = np.vstack([np.nanmean(a, axis=0)] * n)
.....: b_bar = np.vstack([np.nanmean(b, axis=0)] * n)
.....: A = a - a_bar - a_bar.T + np.full(shape=(n, n), fill_value=a_bar.mean())
.....: B = b - b_bar - b_bar.T + np.full(shape=(n, n), fill_value=b_bar.mean())
.....: cov_ab = np.sqrt(np.nansum(A * B)) / n
.....: std_a = np.sqrt(np.sqrt(np.nansum(A ** 2)) / n)
.....: std_b = np.sqrt(np.sqrt(np.nansum(B ** 2)) / n)
.....: return cov_ab / std_a / std_b
.....:
In [220]: df = pd.DataFrame(np.random.normal(size=(100, 3)))
In [221]: df.corr(method=distcorr)
Out[221]:
0 1 2
0 1.000000 0.197613 0.216328
1 0.197613 1.000000 0.208749
2 0.216328 0.208749 1.000000Timedeltas
Timedeltas 文档。
The Timedeltas docs.
In [222]: import datetime
In [223]: s = pd.Series(pd.date_range("2012-1-1", periods=3, freq="D"))
In [224]: s - s.max()
Out[224]:
0 -2 days
1 -1 days
2 0 days
dtype: timedelta64[ns]
In [225]: s.max() - s
Out[225]:
0 2 days
1 1 days
2 0 days
dtype: timedelta64[ns]
In [226]: s - datetime.datetime(2011, 1, 1, 3, 5)
Out[226]:
0 364 days 20:55:00
1 365 days 20:55:00
2 366 days 20:55:00
dtype: timedelta64[ns]
In [227]: s + datetime.timedelta(minutes=5)
Out[227]:
0 2012-01-01 00:05:00
1 2012-01-02 00:05:00
2 2012-01-03 00:05:00
dtype: datetime64[ns]
In [228]: datetime.datetime(2011, 1, 1, 3, 5) - s
Out[228]:
0 -365 days +03:05:00
1 -366 days +03:05:00
2 -367 days +03:05:00
dtype: timedelta64[ns]
In [229]: datetime.timedelta(minutes=5) + s
Out[229]:
0 2012-01-01 00:05:00
1 2012-01-02 00:05:00
2 2012-01-03 00:05:00
dtype: datetime64[ns]In [230]: deltas = pd.Series([datetime.timedelta(days=i) for i in range(3)])
In [231]: df = pd.DataFrame({"A": s, "B": deltas})
In [232]: df
Out[232]:
A B
0 2012-01-01 0 days
1 2012-01-02 1 days
2 2012-01-03 2 days
In [233]: df["New Dates"] = df["A"] + df["B"]
In [234]: df["Delta"] = df["A"] - df["New Dates"]
In [235]: df
Out[235]:
A B New Dates Delta
0 2012-01-01 0 days 2012-01-01 0 days
1 2012-01-02 1 days 2012-01-03 -1 days
2 2012-01-03 2 days 2012-01-05 -2 days
In [236]: df.dtypes
Out[236]:
A datetime64[ns]
B timedelta64[ns]
New Dates datetime64[ns]
Delta timedelta64[ns]
dtype: object与日期时间类似，值可使用 np.nan 设置为 NaT
Values can be set to NaT using np.nan, similar to datetime
In [237]: y = s - s.shift()
In [238]: y
Out[238]:
0 NaT
1 1 days
2 1 days
dtype: timedelta64[ns]
In [239]: y[1] = np.nan
In [240]: y
Out[240]:
0 NaT
1 NaT
2 1 days
dtype: timedelta64[ns]Creating example data
若要从一些给定值的每个组合创建数据框（类似于 R 的 expand.grid() 函数），我们可以创建一个字典，其中键是列名，值是数据值列表：
To create a dataframe from every combination of some given values, like R’s expand.grid() function, we can create a dict where the keys are column names and the values are lists of the data values:
In [241]: def expand_grid(data_dict):
.....: rows = itertools.product(*data_dict.values())
.....: return pd.DataFrame.from_records(rows, columns=data_dict.keys())
.....:
In [242]: df = expand_grid(
.....: {"height": [60, 70], "weight": [100, 140, 180], "sex": ["Male", "Female"]}
.....: )
.....:
In [243]: df
Out[243]:
height weight sex
0 60 100 Male
1 60 100 Female
2 60 140 Male
3 60 140 Female
4 60 180 Male
5 60 180 Female
6 70 100 Male
7 70 100 Female
8 70 140 Male
9 70 140 Female
10 70 180 Male
11 70 180 FemaleConstant series
若要评估序列是否包含常量值，我们可以检查 series.nunique() ⇐ 1。不过，一种性能更高的做法是不先统计所有唯一值，即：
To assess if a series has a constant value, we can check if series.nunique() ⇐ 1. However, a more performant approach, that does not count all unique values first, is:
In [244]: v = s.to_numpy()
In [245]: is_constant = v.shape[0] == 0 or (s[0] == s).all()此方法假设该序列不包含缺失值。对于要删除 NA 值的情况，我们可以先简单地删除这些值：
This approach assumes that the series does not contain missing values. For the case that we would drop NA values, we can simply remove those values first:
In [246]: v = s.dropna().to_numpy()
In [247]: is_constant = v.shape[0] == 0 or (s[0] == s).all()如果缺失值被视为与任何其他值不同，则可以使用：
If missing values are considered distinct from any other value, then one could use:
In [248]: v = s.to_numpy()
In [249]: is_constant = v.shape[0] == 0 or (s[0] == s).all() or not pd.notna(v).any()（请注意，此示例没有消除 np.nan、pd.NA 和 None 之间的歧义）
(Note that this example does not disambiguate between np.nan, pd.NA and None)