我有一个名为鞋码和数量的整数列表。鞋码列表中的整数就像在数量列表中具有数量(在同一索引中)的箱子。我想重新组织列表，使鞋码按升序排列，每个鞋码的所有数量加在一起。
shoesizes = [ 2 , 5 , 6 , 1 , 3 , 2 , 4 , 5 , 2 , 3 , 1 ]
quantities = [ 50, 100, 120, 20, 40, 10, 90 , 10 ,30 , 20, 80]
所以期望的输出将是:
orderedsizes = [ 1 , 2 , 3 , 4 , 5 , 6 ]
totalquant = [100, 90, 60, 90,110, 120]
请您参考如下方法:
因为无论如何你都需要排序，itertools.groupby can be used to do this pretty nicely :
from operator import itemgetter
from itertools import groupby
shoesizes = [ 2 , 5 , 6 , 1 , 3 , 2 , 4 , 5 , 2 , 3 , 1 ]
quantities = [ 50, 100, 120, 20, 40, 10, 90 , 10 ,30 , 20, 80]
# For convenience, short names for itemgetters
getsize, getcnt = itemgetter(0), itemgetter(1)
# Sort to bins of same size next to each other
sorted_bins = sorted(zip(shoesizes, quantities), key=getsize)
# Group and sum bins of same size
summed_sizes = [(k, sum(map(getcnt, g))) for k, g in groupby(sorted_bins, key=getsize)]
# Convert back to separate lists
orderedsizes, totalquant = map(list, zip(*summed_sizes))
print(orderedsizes)
print(totalquant)
哪些输出:
[1, 2, 3, 4, 5, 6]
[100, 90, 60, 90, 110, 120]
也可以用 collections.Counter 来做到这一点(或者只是 collections.defaultdict(int))，然后对 Counter 的 .items() 进行排序；除非 bins 的数量很大，否则性能差异不太重要；如果您一开始就不想排序，Counter 完全避免排序会更有意义。示例代码:
from collections import defaultdict
size_counts = defaultdict(int)
for size, cnt in zip(shoesizes, quantities):
size_counts[size] += cnt
orderedsizes, totalquant = map(list, zip(*sorted(size_counts.items())))
