I have a string with numbers, I can split to individual numbers like
mynum = '123456'.replace(".", "")
[int(i) for i in mynum]
Output:
[1, 2, 3, 4, 5, 6]
I need to split to 2 digits like
[12, 34, 56]
also 3 digits
[123, 456]
2 Answers
One clean approach would use the remainder:
inp = 123456
nums = []
while inp > 0:
nums.insert(0, inp % 100)
inp /= 100
print(nums) # [12, 34, 56]
The above can be easily modified e.g. to use groups of three numbers, instead of two.
3 Comments
hanzgs
This has some useful solution as well
Tim Biegeleisen
@hanzgs That looks like a potential duplicate...you may hammer if you wish.
For two-digit splitting:
[int(mynum[i:i+2]) for i in range(0,len(mynum),2)]
For three-digit splitting:
[int(mynum[i:i+3]) for i in range(0,len(mynum),3)]
So for n-digit splitting:
[int(mynum[i:i+n]) for i in range(0,len(mynum),n)]
2 Comments
Manuel Faysse
You should probably give a quick explanation into why that works ! Note also that your bounds in your first range are probably wrong (you are missing the initial 0 if you want to give a step with no kwargs !). Finally, I honestly don't get why you don't just split the string int(mynum[i:i+2]), and have to ressort to base 10 calculations. Answer can be improved !
hashim
@ManuelFaysse Thanks for pointing out the mistakes. The answer is now edited.
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