使用熊猫中的另一列替换一列中的值的有效方法

Question

How do I replace string values in a dataframe column[1] using list of string values in a different column[2].

Data

          0                       1            2              3
0  3000 20%  dummy1 3000 dummy2 20%  [3000, 20%]  dummy1 dummy2

I want to replace string value in column 1 ie "dummy1 3000 dummy2 20%" using list in column 2 ie "[3000, 20%]". So 3000 and 20% are replaced with ""(empty string) from the string to form 3rd column(Result) ie "dummy1 dummy2"

Code

df = pd.DataFrame([['3000 20%', 'dummy1 3000 dummy2 20%']])
df[2] = df[0].str.split(' ')

def replace_string(x):
    repl_string = str(x[1])
    for key in x[2]:
        repl_string = repl_string.replace(key, '')
    return ' '.join(repl_string.split())

df[3] = df.apply(replace_string, axis=1)

I have currently written the above code, which is slow for large dataframe. How do I improve the efficiency of this code or is there any other way to do this?

Answer 1

Use nested list comprehension:

df = pd.DataFrame([['3000 20%', 'dummy1 a 3000 dummy2 20%'],
                   ['abc 2%', 'klmn 3000 dummy2 2%']])
print (df)
          0                         1
0  3000 20%  dummy1 a 3000 dummy2 20%
1    abc 2%       klmn 3000 dummy2 2%

df[3] = [' '.join(y for y in j.split() if y not in i.split()) for i, j in zip(df[0], df[1])]
print (df)
          0                         1                 3
0  3000 20%  dummy1 a 3000 dummy2 20%   dummy1 a dummy2
1    abc 2%       klmn 3000 dummy2 2%  klmn 3000 dummy2