熊猫：检查数据框的每个单元格中是否存在值

Question

I'm using python 2.7 and want to create a column depending on the existence of each value of a list in every cell.

Here's an example of data:

|    query      |
-----------------
| handbag woman |
| shoe man      |
| t-shirt baby  |
| watch unisex  |
| dress         |

I have a list of value that I want to check:

gender_list=['woman', 'man', 'baby', 'unisex']

the result that I expect:

|   query      |   gender
-----------------------
|   handbag    |  woman 
|   shoe       |  man      
|   t-shirt    |  baby  
|   watch      |  unisex  
|   dress      |  None

Here's what I have already tried:

for gender in gender_list:
    df['gender']=df['query'].map(lambda x : gender if (x.find(gender) != -1) else None)
    df['query']=df['query'].map(lambda x : x.replace(gender, '').strip() if (x.find(gender) != -1) else x)

Answer 1

First in pandas the best is not used loops, because slow (apply are loops under the hood) and rather use vectorized solutions.

Use extract and replace by regex join all values by | and use word boundary for exact match:

gender_list=['woman', 'man', 'baby', 'unisex']
#exact match is not important
#pat = '|'.join(gender_list)
pat = '|'.join(r"\b{}\b".format(x) for x in gender_list)
print (pat)
\bwoman\b|\bman\b|\bbaby\b|\bunisex\b

df['gender'] = df['query'].str.extract('('+ pat + ')', expand=False)
df['query'] = df['query'].str.replace(pat, '').str.strip()
print (df)
     query  gender
0  handbag   woman
1     shoe     man
2  t-shirt    baby
3    watch  unisex
4    dress     NaN

Differences:

print (df)
           query
0  handbag woman
1      shoe many <-man change to many
2   t-shirt baby
3   watch unisex
4          dress

gender_list=['woman', 'man', 'baby', 'unisex']
pat = '|'.join(r"\b{}\b".format(x) for x in gender_list)
df['gender'] = df['query'].str.extract('('+ pat + ')', expand=False)
df['query'] = df['query'].str.replace(pat, '').str.strip()
print (df)
       query  gender
0    handbag   woman
1  shoe many     NaN <-many not extracted
2    t-shirt    baby
3      watch  unisex
4      dress     NaN

---

gender_list=['woman', 'man', 'baby', 'unisex']
pat = '|'.join(gender_list)
df['gender'] = df['query'].str.extract('('+ pat + ')', expand=False)
df['query'] = df['query'].str.replace(pat, '').str.strip()
print (df)
     query  gender
0  handbag   woman
1   shoe y     man <-stay y from many
2  t-shirt    baby
3    watch  unisex
4    dress     NaN

Timings :

df = pd.DataFrame({'query': ['handbag woman', 'shoe man', 't-shirt baby', 'watch unisex', 'dress', 'manpower']})
print (df)

df = pd.concat([df] * 10000, ignore_index=True)


In [299]: %%timeit
     ...: pat = '|'.join(r"\b{}\b".format(x) for x in gender_list)
     ...: df['gender'] = df['query'].str.extract('('+ pat + ')', expand=False)
     ...: df['query'] = df['query'].str.replace(pat, '').str.strip()
     ...: 
     ...: 
1 loop, best of 3: 143 ms per loop

In [300]: %%timeit
     ...: gender_set = set(gender_list)
     ...: 
     ...: def gender_sep(row):
     ...:     lst = row['query'].split(' ')
     ...:     gender = next(iter(gender_set & set(lst)), None)
     ...:     return (' '.join(lst), None) if not gender else \
     ...:            (' '.join(i for i in lst if i!= gender), gender)
     ...: 
     ...: df['query'], df['gender'] = list(zip(*df.apply(gender_sep, axis=1)))
     ...: 
1 loop, best of 3: 933 ms per loop

EDIT:

For more common general solution need escape regex values by re.escape :

import re

gender_list=['woman', 'man', 'baby', 'girl & boy']
pat = '|'.join(r"\b{}\b".format(re.escape(x)) for x in gender_list)
df['gender'] = df['query'].str.extract('('+ pat + ')', expand=False)
df['query'] = df['query'].str.replace(pat, '').str.strip()

Answer 2

This is one way. It's not the most efficient, but it is readable and easy to adapt / maintain.

import pandas as pd

df = pd.DataFrame({'query': ['handbag woman', 'shoe man', 't-shirt baby', 'watch unisex', 'dress', 'manpower']})

gender_list = ['woman', 'man', 'baby', 'unisex']
gender_set = set(gender_list)

def gender_sep(row):
    lst = row['query'].split(' ')
    gender = next(iter(gender_set & set(lst)), None)
    return (' '.join(lst), None) if not gender else \
           (' '.join(i for i in lst if i!= gender), gender)

df['query'], df['gender'] = list(zip(*df.apply(gender_sep, axis=1)))

#       query  gender
# 0   handbag   woman
# 1      shoe     man
# 2   t-shirt    baby
# 3     watch  unisex
# 4     dress    None
# 5  manpower    None