CODEWITHSUNDEEP
pythonarraysstringfor-loop
hafiddle
hafiddle

Reputation: 13

replace the value of 2 arrays in Python String

I have a string and 2 arrays like below:

st="a1b2c3d"
arr1 = ['1','2','3']
arr2 = ['X','Y','Z']

I want to replace all the value of '1', '2', '3' to 'X', 'Y', 'Z'. The final string will look like:

'aXbYcZd'

So I wrote this for loop:

for i in range(0, len(arr1)):
    st.replace(str(arr1[i]),str(arr2[i]))

The result is:

'aXb2c3d'
'a1bYc3d'
'a1b2cZd'

How to correctly do what I want above?

Thanks!

Upvotes: 1

Views: 1221

Answers (3)

Antti Haapala -- Слава Україні
Antti Haapala -- Слава Україні

Reputation: 134066

If you're replacing characters, instead of the inefficient replace loop use str.translate with str.maketrans:

>>> table = str.maketrans('123', 'XYZ')
>>> result = 'a1b2c3d'.translate(table)
>>> result
'aXbYcZd'

maketrans requires 2 strings as arguments. If you really have a list, you can use ''.join(l) to make it into a suitable string. You need to make the table only once.

The efficiency is but one point. str.translate is the way to do this correctly in cases where you will map a => b and b => something else. If you want to replace strings then you might need to use re.sub instead.

Upvotes: 2

John Zwinck
John Zwinck

Reputation: 249642

Calling replace over and over means you have to iterate through the entire string for each replacement, which is O(m * n). Instead:

rep = dict(zip(arr1, arr2)) # make mapping, O(m)
result = ''.join(rep.get(ch, ch) for ch in st)

The first line is O(m), where m is the length of arr1 and arr2.

The second line is O(n), where n is the length of st.

In total this is O(m + n) instead of O(m * n), which is a significant win if either m or n is large.

Upvotes: 0

Austin
Austin

Reputation: 26057

Use zip() to iterate through two lists simultaneously to replace values:

st = "a1b2c3d"
arr1 = ['1','2','3']
arr2 = ['X','Y','Z']

for x, y in zip(arr1, arr2):
    st = st.replace(x, y)

print(st)
# aXbYcZd

str.replace() does not replace a string in-place. You need to assign returned value back to a variable.

Upvotes: 2

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