I want to make replace of all values that are in list(i) for a string (li) and return final string with replaced values.
Instead, my code replaces only one value at a time. Is there a way to fix this simple for loop?
li = str('Text 1. Text 2? Text 3!')
i = ['.', '?', '!']
for y in i:
t = li.replace(y,'')
print(t)
You need to re-assign to li each time, kind of recursive replacement. Because for now you always go from original li to t with a different replacement letter
li = 'Text 1. Text 2? Text 3!'
i = ['.', '?', '!']
for y in i:
li = li.replace(y, '')
You can also use regex module re and pattern [.?!]
li = 'Text 1. Text 2? Text 3!'
i = ['.', '?', '!']
li = re.sub("[" + "".join(i) + "]", '', li)
You have to update the value of li everytime replace method is used
li = 'Text 1. Text 2? Text 3!'
i = ['.', '?', '!']
for y in i:
li = li.replace(y, '')
print(li)
this way the loop checks for the first string to replace then re-assigns the new li value to li and then checks for next string to replace.
The problem is that you are replacing sub strings in li but assigning the returned value to t. In theory, t is always overwritten when you assigned the replaced string to t and li remains the same because you never assigned the replaced string to li
So change the code to this
li = str('Text 1. Text 2? Text 3!')
i = ['.', '?', '!']
for y in i:
li = li.replace(y,'')
print(li)
I think you should use filter in this case, like this:
li = 'Text 1. Text 2? Text 3!'
t = ''.join(list(filter(lambda x: x not in '.?!', li)))
print(t)
U can try this:
li = 'Text 1. Text 2? Text 3!'
i = ['.', '?', '!']
t = li
for y in i:
t = t.replace(y, '')
print(t)
Value of li remains as it was. There is no need to use str function. Output:
Text 1 Text 2 Text 3
