Question

我有一个数据集，在其中我将column1的每个值与column2的所有值进行比较。我能够为每一行创建一个二进制变量，注意是否确实在column2的某处找到了column1值。

我现在想创建一个列，该列是在列2值中找到column1值的所有索引位置的列表。使用Python 3.6

import pandas as pd
import numpy as np

data = [{'column1': 'ibm', 'column2': 'apple'},
    {'column1': 'microsoft', 'column2': 'ibm'},
    {'column1': 'apple', 'column2': 'ibm'},
    {'column1': 'apple', 'column2': 'microsoft'},
    {'column1': 'yahoo', 'column2': 'microsoft'}]

data_df = pd.DataFrame(data)

data_df['match'] = np.where((data_df.column1.isin(data_df['column2'])), 1, 0)

此结果对于该部分是正确的。

   split1     split2      match
0   ibm        apple        1
1   microsoft  ibm          1
2   apple      ibm          1
3   apple      microsoft    1
4   yahoo      microsoft    0

要为column2中找到的column1中的每个值创建索引位置列表，我已经尝试过：

data_df['indices'] = [i for i, x in enumerate(data_df['column2']) if x == np.where((data_df.column1.isin(data_df['column2'])))]

但是，出现以下错误：

data_df['indices'] = [i for i, x in enumerate(data_df['split2']) if x == np.where((data_df.split1.isin(data_df['split2'])))]
Traceback (most recent call last):
  File "<input>", line 1, in <module>
  File "/home/carterrees/PycharmProjects/data_services_predictopotamus/venv_predictopotamus36/lib64/python3.6/site-packages/pandas/core/frame.py", line 3119, in __setitem__
self._set_item(key, value)
  File "/home/carterrees/PycharmProjects/data_services_predictopotamus/venv_predictopotamus36/lib64/python3.6/site-packages/pandas/core/frame.py", line 3194, in _set_item
value = self._sanitize_column(key, value)
  File "/home/carterrees/PycharmProjects/data_services_predictopotamus/venv_predictopotamus36/lib64/python3.6/site-packages/pandas/core/frame.py", line 3391, in _sanitize_column
value = _sanitize_index(value, self.index, copy=False)
  File "/home/carterrees/PycharmProjects/data_services_predictopotamus/venv_predictopotamus36/lib64/python3.6/site-packages/pandas/core/series.py", line 4001, in _sanitize_index
raise ValueError('Length of values does not match length of ' 'index')
ValueError: Length of values does not match length of index

我希望看到的是这个

      split1     split2    match  indices
0      ibm        apple      1     1,2
1      microsoft  ibm        1     3,4
2      apple      ibm        1      0
3      apple      microsoft  1      0
4      yahoo      microsoft  0      Nan

Answer 1

通过首先创建将公司映射到索引的字典，然后通过线性扫描“ column1”简单地查询字典，即可有效地构建“索引”列。

此后，您可以从“索引”派生“匹配”列。

from collections import defaultdict

d = defaultdict(list)
for i, company in enumerate(df['column2']):
    d[company].append(str(i))

d
# defaultdict(list, {'apple': ['0'], 'ibm': ['1', '2'], 'microsoft': ['3', '4']})

# Now comes the fun part.
idx_mapping = {k: ','.join(v) for k, v in d.items()}
df['indices'] = [idx_mapping.get(x, np.nan) for x in df['column1']]
df['match'] = df['indices'].notna()
df

     column1    column2  match indices
0        ibm      apple   True     1,2
1  microsoft        ibm   True     3,4
2      apple        ibm   True       0
3      apple  microsoft   True       0
4      yahoo  microsoft  False     NaN

Answer 2

factorize + stack + np.flatnonzero：

f, l = pd.factorize(df.stack())
r = f.reshape(df.shape)
m = r[:, 0, None] == r[:, 1]

df.assign(
    indices=[np.flatnonzero(c) for c in m],
    match=m.sum(1).astype(bool)
)

     column1    column2 indices  match
0        ibm      apple  [1, 2]   True
1  microsoft        ibm  [3, 4]   True
2      apple        ibm     [0]   True
3      apple  microsoft     [0]   True
4      yahoo  microsoft      []  False

两列的匹配值和索引位置的返回列表

2 个答案: