问题
I have a list of integers, and I want to generate a list containing a list of all the continuous integers.
#I have:
full_list = [0,1,2,3,10,11,12,59]
#I want:
continuous_integers = [[0,1,2,3], [10,11,12], [59]]
I have the following which works, but seems like a poor way to do it:
sub_list = []
continuous_list = []
for x in full_list:
if sub_list == []:
sub_list.append(x)
elif x-1 in sub_list:
sub_list.append(x)
else:
continuous_list.append(sub_list)
sub_list = [x]
continuous_list.append(sub_list)
I've seen other questions suggesting that itertools.groupby is an efficient way to do this, but I'm not familiar with that function and I seem to be having trouble with writing a lambda function to describe the continuous nature.
Question: Is there a better way to be doing this (possibly with itertools.groupby?)
Considerations: full_list will have between 1 and 59 integers, will always be sorted, and integers will be between 0 and 59.
回答1:
You can use the following recipe:
from operator import itemgetter
from itertools import groupby
full_list = [0,1,2,3,10,11,12,59]
cont = [map(itemgetter(1), g) for k, g in groupby(enumerate(full_list), lambda (i,x):i-x)]
# [[0, 1, 2, 3], [10, 11, 12], [59]]
来源:https://stackoverflow.com/questions/15276156/python-return-lists-of-continuous-integers-from-list