Replace String From Tuple Inside List

Question

I currently have the following list:

data = [('b','..','o','b'),('t','s','..','t')]

I am trying to figure out a way to replace all instances of the '..' string to another string. In my instance, the string of ' '.

I have attempted to use the built-in function using the following method, but had no luck.

newData = list(map(lambda i: str.replace(i, ".."," "), data))

Could someone point me in the right direction? My desired output would be the following:

newData = [('b',' ','o','b'),('t','s',' ','t')]

Community · Accepted Answer · 2017-05-23 11:52:47Z

3

You can use a list comprehension with a conditional expression:

>>> data = [('b','..','o','b'),('t','s','..','t')]
>>> newData = [tuple(s if s != ".." else " " for s in tup) for tup in data]
>>> newData
[('b', ' ', 'o', 'b'), ('t', 's', ' ', 't')]
>>>

edited May 23, 2017 at 11:52

CommunityBot

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answered Apr 14, 2014 at 21:05

user2555451

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3 Comments

taytortot Over a year ago

Thanks! This worked flawlessly. I understand where the variable of s comes into play, but could you explain to me the purpose of the "tup" variable?

user2555451 Over a year ago

@taytortot - Certainly. The for tup in data clause will iterate over each item (tuple) inside the list data. It will take each one, store it in the variable tup, and then send it to the tuple(s if s != ".." else " " for s in tup) part. I named the variable tup to represent "tuple".

SethMMorton Over a year ago

@iCodez It might be easier to show what the tup variable is for if you show the equivalent nested for loops.

joc · Accepted Answer · 2014-04-14 21:28:55Z

3

Map operates on each of the elements of the list given. So in this case, your lambda expression is being called on a tuple, not the elements of the tuple as you intended.

wrapping your code in a list comprehension would do it:

newData = [tuple(map(lambda i: str.replace(i, ".."," "), tup)) for tup in data]

answered Apr 14, 2014 at 21:28

joc

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