duplicate list items when using .replace() function

Question

lst = ['123,456', '"hello"', '345,678', '"bye"']
def main():
    new_lst = []
    for item in lst: 
        #print item
        new_lst.append(item.replace(',','***'))
        new_lst.append(item.replace('\"', ''))
    return new_lst

print main()

This is quite puzzling to me. I don't know what I'm doing wrong here. I know it's a very stupid mistake but it's not clicking for me. I don't know why I get an output of:

['123***456', '123,456', '"hello"', 'hello', '345***678', '345,678', '"bye"', 'bye']

What I was hoping was for:

['123***456', 'hello', '345***678', 'bye']

Any help is greatly appreciated!

thefourtheye · Accepted Answer · 2014-09-10 03:10:33Z

2

You are appending the same string twice, with two different replacements. You should chain the replaces like this

new_lst.append(item.replace(',','***').replace('\"', ''))

Even better, you can use a list comprehension here, like this

return [item.replace(',','***').replace('\"', '') for item in lst]

answered Sep 10, 2014 at 3:10

thefourtheye

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2 Comments

jason adams Over a year ago

I didn't know Python could accept two replaces in one statement. Thanks!

Dan Lenski Over a year ago

replace isn't a statement, it's a method of string objects. Since it is a method of string objects and also returns a string, you can chain them together to your heart's content: s.replace(a,b).replace(c,d).replace(x,y)...

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