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python - 如何 'unstack'数据框列中的所有列表并成对组合结果?

转载 作者:太空宇宙 更新时间:2023-11-04 04:40:13 25 4
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我有一个包含 5 列的数据框,每列包含可变长度的列表。这是我的数据框中的一行的样子:

               A                                         B
1 [aircrafts, they, agreement, airplane] [are, built, made, built]

现在我想“拆开”或“拆开”这些列表,以便每个单元格只包含一个值(一个词)。在解包过程中,一列单元格中的单词应与下一列中的相应值成对组合。结果将是:

        A            B
1 aircrafts are
2 they built
3 agreement made
4 airplane built

作为引用,我的完整数据框如下所示:

   obj                       rel1  \
0 [Boeing] [sells]
1 [aircrafts, they, agreement, airplane] [are, built, made, built]
2 [exception, these] [are, are]
3 [sales, contract] [regulated, consist]
4 [contract] [stipulates]
5 [acquisition] [has]
6 [contract] [managed]
7 [employee] [act]
8 [salesperson, Boeing] [change, ensures]
9 [airlines, airlines] [related, have]
10 [Boeing] [keep]

dep1 rel2 \
0 [aircrafts] [to]
1 [] [on, with]
2 [] [of, of, for]
3 [] [by, of, with, of, of]
4 [elements] [across, as]
5 [details] [of, as, of, for]
6 [] [by]
7 [] [as, for]
8 [] [Given, of, over, for]
9 [company] [to, for]
10 [track, aircrafts] [of, between, to, of]

dep2
0 [companies]
1 [demand, customer]
2 [airplanes, scope, case]
3 [means, contracts, companies, acquisitions, ai...
4 [acquisitions, conditions]
5 [airplane, model, airplane, options]
6 [salesperson]
7 [salesperson, contracts]
8 [term, contracts, time, client]
9 [other, example]
10 [relationships, companies, airlines, buyer]

如何在 python 中执行“解包”和重新排列操作?如果这些操作可以在数据帧本身上执行,那就太好了。如果事实证明这很困难或不可能,有没有办法在将列表组合成数据框之前重新排列列表中的数据?

非常感谢您提供的任何帮助或建议。

最佳答案

我相信你希望能够将行读成句子

import pandas as pd
import numpy as np

df = pd.DataFrame(dict(
obj=[['Boeing'], ['aircrafts', 'they', 'agreement', 'airplane'], ['exception', 'these']],
rel1=[['sells'], ['are', 'built', 'made', 'built'],['are', 'are']],
dep1=[['aircrafts'], [], []],
rel2=[['to'], ['on', 'with'], ['of', 'of', 'for']]
))

# Output dataframe
out = pd.DataFrame()

# Keep track of which set of rows we have already appended to the dataframe
row_counter = 0

# Loop through each row in the input dataframe
for row, value in df.iterrows():

# Get the max len of each list in this row
rows_needed = value.map(lambda x: len(x)).max()

for i in range(rows_needed):

# Set a name for this row (numeric)
new_row = pd.Series(name=row_counter+i)

# Loop through each header and create a single row per item in the list
for header in value.index:

# Find if there will be a value here or null
this_value = np.nan
if i < len(df.loc[row, header]):
this_value = df.loc[row, header][i]

# Add this single result to a series for this row
new_row = new_row.set_value(header, this_value)

# Add this row series to the full dataframe
out = out.append(new_row)
row_counter += rows_needed

out
Out[1]:
dep1 obj rel1 rel2
0 aircrafts Boeing sells to
1 NaN aircrafts are on
2 NaN they built with
3 NaN agreement made NaN
4 NaN airplane built NaN
5 NaN exception are of
6 NaN these are of
7 NaN NaN NaN for

order = ['obj', 'rel1', 'dep1', 'rel2']
out = out[order]
out
Out[2]:
obj rel1 dep1 rel2
0 Boeing sells aircrafts to
1 aircrafts are NaN on
2 they built NaN with
3 agreement made NaN NaN
4 airplane built NaN NaN
5 exception are NaN of
6 these are NaN of
7 NaN NaN NaN for

关于python - 如何 'unstack'数据框列中的所有列表并成对组合结果?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50830592/

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