如何匹配来自多个数据框的字符串并使用 AND 和 OR 选项返回索引

Question

这是我想要搜索并取回匹配行号的数据框。'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框。其中每一行都必须搜索为 AND，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

所需的输出[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

了解results = df2.isin( [ term ] )TypeError: unhashable type: 'list'

对于OR，应该很容易购买必须排除已经在第一部分中说明的AND部分

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

Answer 1

这不是您期望的输出，但我要求在 AND 条件中提供索引。生成的输出列表包含 df 逐行的 df 索引。这符合您问题的意图吗？

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]]