根据列值输出多个文件python pandas [英] output multiple files based on column value python pandas

问题描述

我有一个示例熊猫数据框:

i have a sample pandas data frame:

import pandas as pd

df = {'ID': [73, 68,1,94,42,22, 28,70,47, 46,17, 19, 56, 33 ],
  'CloneID': [1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 ],
  'VGene': ['64D', '64D', '64D', 61, 61, 61, 311, 311, 311, 311, 311,  311, 311, 311]}
df = pd.DataFrame(df)

它看起来像这样:

df
Out[7]: 
    CloneID  ID VGene
0         1  73   64D
1         1  68   64D
2         1   1   64D
3         1  94    61
4         1  42    61
5         2  22    61
6         2  28   311
7         3  70   311
8         3  47   311
9         3  46   311
10        4  17   311
11        4  19   311
12        4  56   311
13        4  33   311

我想编写一个简单的脚本，将每个cloneID输出到不同的输出文件.因此，在这种情况下，将有4个不同的文件. 第一个文件将被命名为'CloneID1.txt'，看起来像这样:

i want to write a simple script to output each cloneID to a different output file. so in this case there would be 4 different files. the first file would be named 'CloneID1.txt' and it would look like this:

CloneID  ID   VGene
     1   73   64D
     1   68   64D
     1   1    64D
     1   94   61
     1   42   61

第二个文件将被命名为"CloneID2.txt":

second file would be named 'CloneID2.txt':

CloneID  ID  VGene
     2   22   61
     2   28   311

第三个文件将被命名为"CloneID3.txt":

third file would be named 'CloneID3.txt':

CloneID  ID  VGene
     3   70   311
     3   47   311
     3   46   311

最后一个文件为"CloneID4.txt":

and last file would be 'CloneID4.txt':

CloneID  ID VGene 
    4    17   311
    4    19   311
    4    56   311
    4    33   311

我在网上找到的代码是:

the code i found online was:

import pandas as pd
data = pd.read_excel('data.xlsx')

for group_name, data in data.groupby('CloneID'):
    with open('results.csv', 'a') as f:
        data.to_csv(f)

但是它将所有内容输出到一个文件而不是多个文件.

but it outputs everything to one file instead of multiple files.

推荐答案

您可以执行以下操作:

In [19]:
gp = df.groupby('CloneID')
for g in gp.groups:
    print('CloneID' + str(g) + '.txt')
    print(gp.get_group(g).to_csv())

CloneID1.txt
,CloneID,ID,VGene
0,1,73,64D
1,1,68,64D
2,1,1,64D
3,1,94,61
4,1,42,61

CloneID2.txt
,CloneID,ID,VGene
5,2,22,61
6,2,28,311

CloneID3.txt
,CloneID,ID,VGene
7,3,70,311
8,3,47,311
9,3,46,311

CloneID4.txt
,CloneID,ID,VGene
10,4,17,311
11,4,19,311
12,4,56,311
13,4,33,311

因此，这里我们遍历for g in gp.groups:中的组，并使用它来创建结果文件路径名并在该组上调用to_csv，以便以下内容对您有用:

So here we iterate over the groups in for g in gp.groups: and we use this to create the result file path name and call to_csv on the group so the following should work for you:

gp = df.groupby('CloneID')
for g in gp.groups:
    path = 'CloneID' + str(g) + '.txt'
    gp.get_group(g).to_csv(path)

实际上，以下内容甚至会更简单:

Actually the following would be even simpler:

gp = df.groupby('CloneID')
gp.apply(lambda x: x.to_csv('CloneID' + str(x.name) + '.txt'))

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