replace part of string in columns pandas bases on condition with for loop-CodePudding

I try to format the values in a pandas column based on a condition to a value in another column

wrote this function to format the keys to a brand_specific standard

brand_1 = ['coke', 'pepsi']
def washbox_brand_1(item):
    for item in range(len(df)):
        if df['LEVERANC'][item] in brand_1:
            df['ZOEKCODE'][item] =  df['ZOEKCODE'][item].str.replace('-','')
                  
df2 = df.apply(washbox_brand_1)
df2

Condition works fine but setting the new value is a problem.

gives error:

AttributeError: 'str' object has no attribute 'str'

How can I replace old values to the new format ? Please keep it simple, because code will be extended with much more format_rules for each brand

CodePudding user response：

add condition to check type of data you are replacing, looks like column data is of str type.

if type(df['ZOEKCODE'][item]) == str:
    df['ZOEKCODE'][item] =  df['ZOEKCODE'][item].replace('-','')
else:
    df['ZOEKCODE'][item] =  df['ZOEKCODE'][item].str.replace('-','')

CodePudding user response：

I don't think you need for loop in this case either like @rpanai said in the comments when you are using .apply, unless otherwise you have a specific reason. You could make the function even simpler I think.

Example:

def washbox_brand_1(row):
    brand_1 = ['coke', 'pepsi']
    if row["LEVERANC"] in brand_1:
        row["ZOEKCODE"] = row["ZOEKCODE"].replace('-', '')
    return row
    
df2 = df.apply(washbox_brand_1, axis=1)

You can also use .loc and .isin to get same result. In this case you call only the function without .apply

def washbox_brand_1(df):
    brand_1 = ['coke', 'pepsi']
    df.loc[df["LEVERANC"].isin(brand_1), "ZOEKCODE"] = df.loc[df["LEVERANC"].isin(brand_1), "ZOEKCODE"].str.replace('-', '')
    return df
    
df2 = washbox_brand_1(df)

This should produce the same result and if not, you will have to provide a minimal reproducible example just like @rpanai made mentioned in the comment section.