CODEWITHSUNDEEP
listpython-3.xslice
tktktk0711
tktktk0711

Reputation: 1694

Python3: how to slice a list to two part in a loop, one is 20%, the left is another part

Hi I want to slice a large list data like the following list, one part is 20%(consecutive) part of the list, another is left. I just know to get the 20% part but how to get the left part of list in a loop. Thanks my code is following. for example, when X1 = [ '1', '2'], X2 = ['3','4','5','6', 7','8','9','10']. then X1 = ['2','3'], and X2 = ['1','4','5','6', 7','8','9','10'], .........

X = ['1','2','3','4','5','6', 7','8','9','10']
index = 0
for i in range(9):
     X1 = X[i:(i+2)]
     X2 = X[]

Upvotes: 0

Views: 54

Answers (2)

cegas
cegas

Reputation: 3091

I would like to offer a similar, but more robust solution. You say you need to get 20% of the large list. While slice size of 2 works for a list with 9 members, you will obviously need to increase it for a larger list. If you do not know how many members a list will have, it can be particularly tedious and error prone. Hence, I would suggest something like this:

import math

X = ['1','2','3','4','5','7','8','9','10']
# Calculate how many members a 20% slice should have
twenty = int(math.ceil(len(X) *  20 / 100.0))

for i in range(len(X)):
     X1 = X[i:(i+twenty)]
     X2 = X[:i] + X[(i+twenty):]

What this does is calculate how many members ~20% slice should have, and then apply this number to list slicing. Note, however, that math.ceil() always rounds up, so if you have something like 2.1, it would end up being 3.0 (meaning that some slices will have slightly more members than 20%, e.g. 22%). If you want to round in a more familiar fashion (2.49 -> 2; 2.5 -> 3), round() would be a good choice.

Upvotes: 1

omri_saadon
omri_saadon

Reputation: 10631

You can fetch the second part by telling from what index you would like to start untill the end.

X = ['1','2','3','4','5','7','8','9','10']
index = 0
for i in range(9):
     X1 = X[i:(i+2)]
     X2 = X[:i] + X[(i+2):]

Upvotes: 1

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