Python 获取多个 Pandas DataFrame 的平均值
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Get the mean across multiple Pandas DataFrames
提问by Tim
I'm generating a number of dataframes with the same shape, and I want to compare them to one another. I want to be able to get the mean and median across the dataframes.
我正在生成许多具有相同形状的数据框，我想将它们相互比较。我希望能够获得跨数据帧的平均值和中位数。
Source.0 Source.1 Source.2 Source.3
cluster
0 0.001182 0.184535 0.814230 0.000054
1 0.000001 0.160490 0.839508 0.000001
2 0.000001 0.173829 0.826114 0.000055
3 0.000432 0.180065 0.819502 0.000001
4 0.000152 0.157041 0.842694 0.000113
5 0.000183 0.174142 0.825674 0.000001
6 0.000001 0.151556 0.848405 0.000038
7 0.000771 0.177583 0.821645 0.000001
8 0.000001 0.202059 0.797939 0.000001
9 0.000025 0.189537 0.810410 0.000028
10 0.006142 0.003041 0.493912 0.496905
11 0.003739 0.002367 0.514216 0.479678
12 0.002334 0.001517 0.529041 0.467108
13 0.003458 0.000001 0.532265 0.464276
14 0.000405 0.005655 0.527576 0.466364
15 0.002557 0.003233 0.507954 0.486256
16 0.004161 0.000001 0.491271 0.504568
17 0.001364 0.001330 0.528311 0.468996
18 0.002886 0.000001 0.506392 0.490721
19 0.001823 0.002498 0.509620 0.486059
Source.0 Source.1 Source.2 Source.3
cluster
0 0.000001 0.197108 0.802495 0.000396
1 0.000001 0.157860 0.842076 0.000063
2 0.094956 0.203057 0.701662 0.000325
3 0.000001 0.181948 0.817841 0.000210
4 0.000003 0.169680 0.830316 0.000001
5 0.000362 0.177194 0.822443 0.000001
6 0.000001 0.146807 0.852924 0.000268
7 0.001087 0.178994 0.819564 0.000354
8 0.000001 0.202182 0.797333 0.000485
9 0.000348 0.181399 0.818252 0.000001
10 0.003050 0.000247 0.506777 0.489926
11 0.004420 0.000001 0.513927 0.481652
12 0.006488 0.001396 0.527197 0.464919
13 0.001510 0.000001 0.525987 0.472502
14 0.000001 0.000001 0.520737 0.479261
15 0.000001 0.001765 0.515658 0.482575
16 0.000001 0.000001 0.492550 0.507448
17 0.002855 0.000199 0.526535 0.470411
18 0.000001 0.001952 0.498303 0.499744
19 0.001232 0.000001 0.506612 0.492155
Then I want to get the mean of these two dataframes.
然后我想得到这两个数据帧的平均值。
What is the easiest way to do this?
什么是最简单的方法来做到这一点？
Just to clarify I want to get the mean for each particular cell when the indexes and columns of all the dataframes are exactly the same.
只是为了澄清我想在所有数据帧的索引和列完全相同时获得每个特定单元格的平均值。
So in the example I gave, the average for [0,Source.0]would be (0.001182 + 0.000001) / 2 = 0.0005915.
所以在我给出的例子中，平均值为[0,Source.0](0.001182 + 0.000001) / 2 = 0.0005915。
采纳答案by ali_m
Assuming the two dataframes have the same columns, you could just concatenate them and compute your summary stats on the concatenated frames:
假设两个数据帧具有相同的列，您可以将它们连接起来并在连接的帧上计算您的汇总统计信息：
import numpy as np
import pandas as pd
# some random data frames
df1 = pd.DataFrame(dict(x=np.random.randn(100), y=np.random.randint(0, 5, 100)))
df2 = pd.DataFrame(dict(x=np.random.randn(100), y=np.random.randint(0, 5, 100)))
# concatenate them
df_concat = pd.concat((df1, df2))
print df_concat.mean()
# x -0.163044
# y 2.120000
# dtype: float64
print df_concat.median()
# x -0.192037
# y 2.000000
# dtype: float64
Update
更新
If you want to compute stats across each set of rows with the same index in the two datasets, you can use .groupby()to group the data by row index, then apply the mean, median etc.:
如果要计算两个数据集中具有相同索引的每组行的统计数据，可以使用.groupby()按行索引对数据进行分组，然后应用平均值、中位数等：
by_row_index = df_concat.groupby(df_concat.index)
df_means = by_row_index.mean()
print df_means.head()
# x y
# 0 -0.850794 1.5
# 1 0.159038 1.5
# 2 0.083278 1.0
# 3 -0.540336 0.5
# 4 0.390954 3.5
This method will work even when your dataframes have unequal numbers of rows - if a particular row index is missing in one of the two dataframes, the mean/median will be computed on the single existing row.
即使您的数据帧具有不相等的行数，此方法也将起作用 - 如果两个数据帧之一中缺少特定的行索引，则将在单个现有行上计算平均值/中值。
回答by Phillip Cloud
You can simply assign a label to each frame, call it groupand then concatand groupbyto do what you want:
您可以将标签简单地分配给每一帧，调用它group，然后concat与groupby做你想做什么：
In [57]: df = DataFrame(np.random.randn(10, 4), columns=list('abcd'))
In [58]: df2 = df.copy()
In [59]: dfs = [df, df2]
In [60]: df
Out[60]:
a b c d
0 0.1959 0.1260 0.1464 0.1631
1 0.9344 -1.8154 1.4529 -0.6334
2 0.0390 0.4810 1.1779 -1.1799
3 0.3542 0.3819 -2.0895 0.8877
4 -2.2898 -1.0585 0.8083 -0.2126
5 0.3727 -0.6867 -1.3440 -1.4849
6 -1.1785 0.0885 1.0945 -1.6271
7 -1.7169 0.3760 -1.4078 0.8994
8 0.0508 0.4891 0.0274 -0.6369
9 -0.7019 1.0425 -0.5476 -0.5143
In [61]: for i, d in enumerate(dfs):
....: d['group'] = i
....:
In [62]: dfs[0]
Out[62]:
a b c d group
0 0.1959 0.1260 0.1464 0.1631 0
1 0.9344 -1.8154 1.4529 -0.6334 0
2 0.0390 0.4810 1.1779 -1.1799 0
3 0.3542 0.3819 -2.0895 0.8877 0
4 -2.2898 -1.0585 0.8083 -0.2126 0
5 0.3727 -0.6867 -1.3440 -1.4849 0
6 -1.1785 0.0885 1.0945 -1.6271 0
7 -1.7169 0.3760 -1.4078 0.8994 0
8 0.0508 0.4891 0.0274 -0.6369 0
9 -0.7019 1.0425 -0.5476 -0.5143 0
In [63]: final = pd.concat(dfs, ignore_index=True)
In [64]: final
Out[64]:
a b c d group
0 0.1959 0.1260 0.1464 0.1631 0
1 0.9344 -1.8154 1.4529 -0.6334 0
2 0.0390 0.4810 1.1779 -1.1799 0
3 0.3542 0.3819 -2.0895 0.8877 0
4 -2.2898 -1.0585 0.8083 -0.2126 0
5 0.3727 -0.6867 -1.3440 -1.4849 0
6 -1.1785 0.0885 1.0945 -1.6271 0
.. ... ... ... ... ...
13 0.3542 0.3819 -2.0895 0.8877 1
14 -2.2898 -1.0585 0.8083 -0.2126 1
15 0.3727 -0.6867 -1.3440 -1.4849 1
16 -1.1785 0.0885 1.0945 -1.6271 1
17 -1.7169 0.3760 -1.4078 0.8994 1
18 0.0508 0.4891 0.0274 -0.6369 1
19 -0.7019 1.0425 -0.5476 -0.5143 1
[20 rows x 5 columns]
In [65]: final.groupby('group').mean()
Out[65]:
a b c d
group
0 -0.394 -0.0576 -0.0682 -0.4339
1 -0.394 -0.0576 -0.0682 -0.4339
Here, each groupis the same, but that's only because df == df2.
在这里，每个group都是相同的，但这只是因为df == df2.
Alternatively, you can throw the frames into a Panel:
或者，您可以将帧放入一个Panel:
In [69]: df = DataFrame(np.random.randn(10, 4), columns=list('abcd'))
In [70]: df2 = DataFrame(np.random.randn(10, 4), columns=list('abcd'))
In [71]: panel = pd.Panel({0: df, 1: df2})
In [72]: panel
Out[72]:
<class 'pandas.core.panel.Panel'>
Dimensions: 2 (items) x 10 (major_axis) x 4 (minor_axis)
Items axis: 0 to 1
Major_axis axis: 0 to 9
Minor_axis axis: a to d
In [73]: panel.mean()
Out[73]:
0 1
a 0.3839 0.2956
b 0.1855 -0.3164
c -0.1167 -0.0627
d -0.2338 -0.0450
回答by FooBar
I go similar as @ali_m, but since you want one mean per row-column combination, I conclude differently:
我与@ali_m 类似，但由于您想要每行列组合一个均值，因此我得出不同的结论：
df1 = pd.DataFrame(dict(x=np.random.randn(100), y=np.random.randint(0, 5, 100)))
df2 = pd.DataFrame(dict(x=np.random.randn(100), y=np.random.randint(0, 5, 100)))
df = pd.concat([df1, df2])
foo = df.groupby(level=1).mean()
foo.head()
x y
0 0.841282 2.5
1 0.716749 1.0
2 -0.551903 2.5
3 1.240736 1.5
4 1.227109 2.0
回答by ZJS
Here is a solution first unstack both dataframes so they are series with multiindexes(cluster, colnames)... then you can use Series addition and division, which automattically do the operation on the indexes, finally unstack them... here it is in code...
这是一个解决方案，首先解开两个数据帧，使它们与多索引（集群，列名）串联......然后你可以使用系列加法和除法，它会自动对索引进行操作，最后解开它们......这里是代码...
averages = (df1.stack()+df2.stack())/2
averages = averages.unstack()
And your done...
你完成了...
Or for more general purposes...
或者用于更一般的目的...
dfs = [df1,df2]
averages = pd.concat([each.stack() for each in dfs],axis=1)\
.apply(lambda x:x.mean(),axis=1)\
.unstack()
回答by user394430
As per Niklas' comment, the solution to the question is panel.mean(axis=0).
根据 Niklas 的评论，问题的解决方案是panel.mean(axis=0).
As a more complete example:
作为一个更完整的例子：
import pandas as pd
import numpy as np
dfs = {}
nrows = 4
ncols = 3
for i in range(4):
dfs[i] = pd.DataFrame(np.arange(i, nrows*ncols+i).reshape(nrows, ncols),
columns=list('abc'))
print('DF{i}:\n{df}\n'.format(i=i, df=dfs[i]))
panel = pd.Panel(dfs)
print('Mean of stacked DFs:\n{df}'.format(df=panel.mean(axis=0)))
Will give the following output:
将给出以下输出：
DF0:
a b c
0 0 1 2
1 3 4 5
2 6 7 8
3 9 10 11
DF1:
a b c
0 1 2 3
1 4 5 6
2 7 8 9
3 10 11 12
DF2:
a b c
0 2 3 4
1 5 6 7
2 8 9 10
3 11 12 13
DF3:
a b c
0 3 4 5
1 6 7 8
2 9 10 11
3 12 13 14
Mean of stacked DFs:
a b c
0 1.5 2.5 3.5
1 4.5 5.5 6.5
2 7.5 8.5 9.5
3 10.5 11.5 12.5
