How to convert the strings in sublists into integer faster without forloop in python?

Question

I want to convert the strings in sublists into integer. Here is the code,

Example Input:

totalclient=[['5', '110'], ['9', '500'], ['20', '400']]
totalhouse=[['10', '100'], ['2', '200'], ['30', '300']]

This example list has only 4 sublists, But I want to process over 700 sublists for two lists and every sublist will have more than 300 strings!

Code:

totalclient = [list(map(int, sublist)) for sublist in totalclient]
totalhouse = [list(map(int, sublist)) for sublist in totalhouse]

Output

[[5, 110], [9, 500], [20, 400]]
[[10, 100], [2, 200], [30, 300]]

Above code works for me! But for loop takes too much time to convert it is there any other way to convert it into integer without using for loop?

I tried to convert sublist while inputting sublists like belowcode but it also takes same time!

for i in range(0,n[0]):
    client=[int(x) for x in input().split()]
    totalclient.append(client)

You have a use a *for* somewhere, even if you're not typing `for ...` explicitly. Except of course, you want to manually remove the quotes yourself, then no need for *for*. — Moses Koledoye, Oct 17 '17 at 13:17
The above commenter is right, but it might be faster to do it in place? Loop over it and explicitly change each one rather than creating a new list. It's still a for loop, but you avoid needing to create extra space in memory. — James, Oct 17 '17 at 13:19
You guys saying to put like this right? for i in range(0,n[0]): client=[int(x) for x in input().split()] totalclient.append(client) — AzizStark, Oct 17 '17 at 13:21
@AzizStark Your code is fine. Some tasks simply take some time. And how much time are we talking about here? It does not feel like this should be more than a few seconds if even.. — Ma0, Oct 17 '17 at 13:23
Don't you have to iterate through your sublist object before applying it to `map` to convert it into integer? I believe you have not posted the complete code? — Prasad, Oct 17 '17 at 13:25
Yes, i didnot posted the complete code because it is a part of my Bipartite matching Script! — AzizStark, Oct 17 '17 at 13:28
In the end, you still have to convert all the strings to integers. It does not matter how you loop over the collection, there are no shortcuts, unfortunately. — Kendas, Oct 17 '17 at 13:28
possible duplicate: https://stackoverflow.com/q/46341628/7692463 — scharette, Oct 17 '17 at 13:29
This code converts the list elements into strings without for loop! n = list(map(int, n)) — AzizStark, Oct 17 '17 at 13:29
list() is a for loop. It loops over the stuff in the brackets. — Sekuraz, Oct 17 '17 at 13:32
`n = list(map(int, n))` is essentially identical to `n = [int(item) for item in n]`. The fact that you didn't write `for`, does not mean that a for loop doesn't exist. — Kendas, Oct 17 '17 at 13:32
Try collecting your input as integers to begin with, if you really are that concerned about it — OneCricketeer, Oct 17 '17 at 13:34
@cricket_007 As i said that also makes use of for loop and takes same time. — AzizStark, Oct 17 '17 at 13:35
For loop where? No loops are needed for `totalclient=[[5, 110]]`... You're focusing too much on the implementation than benchmarking where the actual bottleneck of the code is — OneCricketeer, Oct 17 '17 at 13:38
If you'd like someone to review your working code, please see https://codereview.stackexchange.com/tour — OneCricketeer, Oct 17 '17 at 13:40
Here is my entire code : http://textuploader.com/d4uzo Could any one help me? — AzizStark, Oct 17 '17 at 13:41
I used pypy3 instead of python it worked 3 times faster than python3!! Python 3 needs to be improved! — AzizStark, Oct 17 '17 at 14:29
Sad that people that people don't upvote the comment though because there will be no accepted/useful answer to the question. — scharette, Oct 17 '17 at 14:34

00sdf0 · Accepted Answer · 2017-10-18T05:53:11.273

This was faster on my machine:

tc = np.array(list(repeat(['1000000000000000','200000000000'], 700)))
tl = np.array(list(repeat(['2000000000000000','300000000000'], 700)))

def s1(tc,tl):
    totalc = [list(map(int, sublist)) for sublist in tc]
    totalh = [list(map(int, sublist)) for sublist in tl]

def s2(tc,tl):
    ntc = list(map(lambda w: [map(int, w)], chain(tc)))
    ntc = list(map(lambda w: [map(int, w)], chain(tl)))

def wrapper(func, *args, **kwargs):
    def wrapped():
        return func(*args, **kwargs)
    return wrapped

wrp1 = wrapper(s1, tc, tl)
wrp2 = wrapper(s2, tc, tl)
print(timeit.timeit(wrp1, number=1000))
print(timeit.timeit(wrp2, number=1000))

>>> wrp1 = 5.529523203999929
>>> wrp2 = 0.8688932149999573

score -1 · Answer 2 · answered Oct 17 '17 at 13:51

-1

try like this,

from functools import *
totalclient=[['5', '110'], ['9', '500'], ['20', '400']]
totalhouse=[['10', '100'], ['2', '200'], ['30', '300']]
a1 = []
a2 = []
for i in totalclient:
    aa = map((lambda x:int(x)),i)
    a1.append(list(aa))
totalclient = a1
for i in totalhouse:
    aa = map((lambda x:int(x)),i)
    a2.append(list(aa))
totalhouse = a2

print(totalclient)
print(totalhouse)

answered Oct 17 '17 at 13:51

vishal paalakurthi

138
1
9

Thanks for the help! but it takes extra time than before! – AzizStark Oct 17 '17 at 14:00

How to convert the strings in sublists into integer faster without forloop in python?

2 Answers2