using a for loop within a .apply() for pandas dataframes

Question

I'd like to remove all items within a list. I've tried the below code to iterate through all items in a list within the context of a pandas .apply(). However, the function remove(x) only seems to be going to the first item in remove_l. How can I make sure it goes through all items in remove_l?

I know I can just create separate if statements, which I've already done, but I'd like to implement using a for loop just in case the list gets longer.

remove_l = [r'[A-Za-z]+(?:\/)', r'Today, ', '-']

def remove(x):
    for phrase in remove_l:
        if re.search(phrase, x):
            if phrase == '-':
                new = x.replace(phrase, ' ')
            else: 
                new = x[re.search(phrase, x).span()[1]:].strip()
            return new 
        else: 
            return x


#check up on items 
#60, 330, 347, 411, 647
#idx = nocountries_df[nocountries_df.Name.str.contains('\/')].Name.index
nocountries_df.Name.apply(lambda x: remove(x))

Andy Hayden · Accepted Answer · 2014-03-18 20:33:09Z

2

This is an indentation problem, when it hits the first return (in the for loop) it returns that value:

def remove(x):
    for phrase in remove_l:
        if re.search(phrase, x):
            if phrase == '-':
                new = x.replace(phrase, ' ')
            else: 
                new = x[re.search(phrase, x).span()[1]:].strip()
            return new  # <- returns here (in first phase) 
        else: 
            return x  # <- or returns here (in first phase)

You want to return after the for loop, it's probably easiest just to change x in the for loop:

def remove(x):
    for phrase in remove_l:
        if re.search(phrase, x):
            if phrase == '-':
                x = x.replace(phrase, ' ')
            else: 
                x = x[re.search(phrase, x).span()[1]:].strip()
    return x

answered Mar 18, 2014 at 20:33

Andy Hayden

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3 Comments

user3314418 Over a year ago

Hm, small problem. I wanted to return x so I can return the dataframe with the same size. how would i do that? I'm noticing that if I incorporate my "else: return x" in there, the items don't change. actually i think i solved it by using 'break'

user3314418 Over a year ago

Hm, actually using 'break' creates a problem because it doesn't go through the entire list... any thoughts?

Andy Hayden Over a year ago

@user3314418 I'm not sure I follow: why doesn't this solution work? It may help to give an input and expected output.

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