Applying a function to a dataframe does not work-CodePudding

Now I have a data frame with entries like '7M'as a string where M is Million and I want to change it to 7000000 This is a look from the data frame

Inpute Example

I have tried to make a function to separate the numbers from the letter and changing the letter to a number and it worked

def num_repair(x):
    if 'B' in x:
        l = 10**9
        x = x[:-1]
        x = pd.to_numeric(x)
        x = x * l
    elif 'TR' in x:
        l = 10**12
        x =x[:-2]
        x = pd.to_numeric(x)
        x = x * l
    elif 'M' in x:
        l = 10**6
        x = x[:-1]
        x = pd.to_numeric(x)
        x = x * l
    return(x)

and when I tried to apply it to the data frame it didn't give me anything any help please?

CodePudding user response：

TL;DR:

What you are looking for is .applymap()

Details:

Your method is actually written well and can be used in .apply() as-is, for a pandas.Series object, but I assume that if you are experiencing issues, it is due to the fact that you are probably using it for a pandas.DataFrame, against multiple columns. In such a case, the argument passed to num_repair is actually of type pandas.Series, which num_repair is not really meant to support. I can only assume, since the code that uses num_repair isn't given. Consider adding it for the completeness of the question.

If so, you can use it as follows:

df = pd.DataFrame([
    ['1M', '1B', '1TR'],
    ['22M', '22B', '22TR'],
], columns=[1990, 1991, 1992])
df.applymap(num_repair)

output:


        1990      1991          1992
0       1000000   1000000000    1000000000000
1       22000000  22000000000   22000000000000

Side Note

If you want to apply it to all columns except the country, since the name may contain B / TR / M - you can do the following:

df = pd.DataFrame([
    ['countryM', '1M', '1B', '1TR'],
    ['countryB', '22M', '22B', '22TR'],
], columns=['country', 1990, 1991, 1992])
df.loc[:, df.columns.drop('country')] = df.loc[:, df.columns.drop('country')].applymap(num_repair)
df

output:

    country     1990        1991        1992
0   countryM    1000000     1000000000  1000000000000
1   countryB    22000000    22000000000 22000000000000

CodePudding user response：

Works fine on my toy example:

>>> df = pd.DataFrame({'a': ['32M', '13B', '33TR']})
>>> df['a'].apply(num_repair)

0          32000000
1       13000000000
2    33000000000000