如果子字符串在列表中，则保留子字符串

Question

I have a column that contains strings and a list that contains strings that I wish to preserve in the column. If a substring is not present in the list, remove it. Note that there is to be no double whitespaces or whitespace at the beginning or end of the string in the column.

How can I accomplish this efficiently?

df['column']
>>>
0    good day happy night
1    good bird sad day
2    day over ready

ls = ['good', 'day']

Output:

df['column']
>>>
0    good day
1    good day
2    day

Answer 1

Use Series.str.findall with joined ls by | for regex OR with Series.str.join for join lists:

ls = ['good', 'day']

df['column'] = df['column'].str.findall('|'.join(ls)).str.join(' ')
print (df)
     column
0  good day
1  good day
2       day

If need match values between space by word boundaries:

#daytime changed
print (df)
                     column
0  good daytime happy night 
1         good bird sad day
2            day over ready

---

ls = ['good', 'day']

pat = '|'.join(r"\b{}\b".format(x) for x in ls)
df['column1'] = df['column'].str.findall(pat).str.join(' ')
df['column2'] = df['column'].str.findall('|'.join(ls)).str.join(' ')
print (df)
                     column   column1   column2
0  good daytime happy night      good  good day
1         good bird sad day  good day  good day
2            day over ready       day       day

---

Another idea is use lambda function and lookup to list converted to sets:

sets = set(ls)
df['column1'] = df['column'].apply(lambda x: ' '.join(y for y in x.split() if y in sets))

Answer 2

With apply and a lambda function:

df['column'].apply(lambda row: ' '.join(list(set(row.split()) & ls)))

Note that by using sets the order of the strings might be changed.

MCVE:

df = pd.DataFrame({'column':['good day happy night', 'good bird sad day', 'day over ready']}) 
#                  column
# 0  good day happy night
# 1     good bird sad day
# 2        day over ready 
ls = ['good', 'day']
ls = set(ls)                                               
df['column'].apply(lambda row: ' '.join(list(set(row.split()) & ls)))
# 0    good day
# 1    good day
# 2         day
# Name: column, dtype: object