Python 如何匹配来自多个数据帧的字符串并使用and和OR选项返回索引

这是我要搜索的数据帧，并返回匹配的行号。

'A'

和

'AB'

是完全不同的东西

df2 = pd.DataFrame(np.array(['A','B','AC','AD','NAN','XX','BC','SLK','AC','AD','NAN','XU','BB','FG','XZ','XY','AD','NAN','NF','XY','AB','AC','AD','NAN','XY','LK','AC','AC','AD','NAN','KH','BC','GF','BC','AD']).reshape(5,7),columns=['a','b','c','d','e','f','g'])


    a   b   c   d   e   f   g
0   A   B   AC  AD  NAN XX  BC
1   SLK AC  AD  NAN XU  BB  FG
2   XZ  XY  AD  NAN NF  XY  AB
3   AC  AD  NAN XY  LK  AC  AC
4   AD  NAN KH  BC  GF  BC  AD

我将搜索的字符串来自这个较小的数据帧。其中每一行都必须作为和进行搜索，以获取数据帧df2的匹配字符串行索引

df = pd.DataFrame(np.array(['A','B','C','D','AA','AB','AC','AD','NAN','BB','BC','AD']).reshape(6,2),columns=['a1','b1'])


a1  b1
0   A   B  # present in the first row of df2
1   C   D  # not present in any row of df2
2   AA  AB # not present in any row of df2
3   AC  AD # present in the second row of df2
4   NAN BB # present in the second row of df2
5   BC  AD # present in the fourth row of df2

和部分

所需输出

[0,1,3,4]

import pandas as pd
import numpy as np


index1 = df.index # Finds the number of row in df
terms=[]
React=[]
for i in range(len(index1)): #for loop to search each row of df dataframe
  terms=df.iloc[i] # Get i row
  terms[i]=terms.values.tolist() # converts to a list
  print(terms[i]) # to check
    # each row
  for term in terms[i]: # to search for each string in the 
    print(term)
    results = pd.DataFrame()
    if results.empty:
      results = df2.isin( [ term ] )
    else:
      results |= df2.isin( [ term ] ) 
  results['count'] = results.sum(axis=1)
  print(results['count'])
  print(results[results['count']==len(terms[i])].index.tolist())
  React=results[results['count']==len(terms[i])].index.tolist()
  React

获取

TypeError:unhable类型：
results=df2.isin（[term]）

对于或应该很容易购买，必须排除已在第一节中说明的零部件

React2=df2.isin([X]).any(1).index.tolist()
React2

这不是您期望的输出，但我要求在AND条件下使用索引。生成的输出列表逐行包含df2索引。这符合你问题的意图吗

output = []
for i in range(len(df)):
    tmp = []
    for k in range(len(df2)):
        d = df2.loc[k].isin(df.loc[i,['a1']])
        f = df2.loc[k].isin(df.loc[i,['b1']])
        d = d.tolist()
        f = f.tolist()
        if sum(d) >= 1 and sum(f) >=1:
            tmp.append(k)
    output.append(tmp)

output
[[0], [], [], [0, 1, 3], [1], [0, 4]]

提高你所谓的“df2”的期望输出。@r初学者非常感谢你的评论。我确实在你的评论后添加了所需的输出。完美这是可行的，但我需要一些时间来测试我的原始数据。请允许我有时12小时就足够我来测试这个了。非常感谢你。