检查字符串是否在另一列熊猫中

Question

Below is my DF

df= pd.DataFrame({'col1': ['[7]', '[30]', '[0]', '[7]'], 'col2': ['[0%, 7%]', '[30%]', '[30%, 7%]', '[7%]']})

col1    col2    
[7]     [0%, 7%]
[30]    [30%]
[0]     [30%, 7%]
[7]     [7%]

The aim is to check if col1 value is contained in col2 below is what I've tried

df['test'] = df.apply(lambda x: str(x.col1) in str(x.col2), axis=1)

Below is the expected output

col1    col2       col3
[7]     [0%, 7%]   True
[30]    [30%]      True
[0]     [30%, 7%]  False
[7]     [7%]       True

Answer 1

Use Series.str.extractall for get numbers, reshape by Series.unstack , so possible compare by DataFrame.isin with DataFrame.any :

df['test'] = (df['col2'].str.extractall('(\d+)')[0].unstack()
                        .isin(df['col1'].str.strip('[]'))
                        .any(axis=1))
print (df)
   col1       col2   test
0   [7]   [0%, 7%]   True
1  [30]      [30%]   True
2   [0]  [30%, 7%]  False
3   [7]       [7%]   True

Answer 2

You can extract the numbers on both columns and join , then check if there is at least one match per id using eval + groupby + any :

(df['col2'].str.extractall('(?P<col2>\d+)').droplevel(1)
   .join(df['col1'].str[1:-1])
   .eval('col2 == col1')
   .groupby(level=0).any()
)

output:

0     True
1     True
2    False
3     True

Answer 3

One approach:

import ast

# convert to integer list
col2_lst = df["col2"].str.replace("%", "").apply(ast.literal_eval)

# check list containment
df["col3"] = [all(bi in a for bi in b)  for a, b in zip(col2_lst, df["col1"].apply( ast.literal_eval)) ]

print(df)

Output

   col1       col2   col3
0   [7]   [0%, 7%]   True
1  [30]      [30%]   True
2   [0]  [30%, 7%]  False
3   [7]       [7%]   True

Answer 4

You can also replace the square brackets with word boundaries \\b and use re.search like in

import re
#...
df.apply(lambda x: bool(re.search(x['col1'].replace("[",r"\b").replace("]",r"\b"), x['col2'])), axis=1)
# => 0     True
#    1     True
#    2    False
#    3     True
#    dtype: bool

This will work because \\b7\\b will find a match in [0%, 7%] as 7 is neither preceded nor followed with letters, digits or underscores. There won't be any match found in [30%, 7%] as \\b0\\b does not match a zero after a digit (here, 3 ).