在数据框上按行将浮点类型数转换为百分比 - Python

Question

Hope you are doing ok. I have the following DataFrame:

Date	Income_type	Mike	Joan
2021/10/31	Salary	25	32
2021/10/31	Investments	10	9
2021/10/31	Investments/Salary	0,4	0,28
2021/09/30	Salary	30	36
2021/09/30	Investments	15	6
2021/09/30	Investments/Salary	0,5	0,16

And I want to turn the Investments /Salary row numbers into a percentage, like this:

Date	Income_type	Mike	Joan
2021/10/31	Salary	25	32
2021/10/31	Investments	10	9
2021/10/31	Investments/Salary	40%	28%
2021/09/30	Salary	30	36
2021/09/30	Investments	15	6
2021/09/30	Investments/Salary	50%	16%

I have tried the following but it hasn't worked:

df['Mike'] = np.where(df['Income_type']=='Investments/salary',df['Mike'].astype(float).map(lambda n: '{:.2%}'.format(n)),df['Mike'])

Any ideas?

Answer 1

Try this:

df.update(
    df.loc[df.Income_type =='Investments/Salary'][['Mike','Jordan']]\
        .replace(to_replace =',', value = '.', regex = True).astype(float)
        )

df.update(df.loc[df.Income_type =='Investments/Salary']['Mike'].map("{:.2%}".format))
df.update(df.loc[df.Income_type =='Investments/Salary']['Jordan'].map("{:.2%}".format))

Answer 2

You can use lambda functions like this. I tested it with synthetic data:

df.Mike = df.Mike.astype(float)

def make_percentage(x):
    if  float(x)<1:
        return str(float(x)*100)+'%'
    else: return x
df.Mike = df.Mike.apply(lambda x: make_percentage(x))