关于 Python 元组操作

Question

我有一个元组列表，每个元组都有一个字符串和一个整数值：

list_a= [('0', 60), ('0', 5), ('2', 1), ('3', 14), ('4', 39), ('1', 17), ('2', 14), ('2', 29), ('4', 1), ('3', 1), ('3', 16), ('4', 8)]

如果字符串相同且仅当字符串彼此相邻时，我想添加值，使输出看起来像：

result=[('0', 65), ('2', 1), ('3', 14), ('4', 39), ('1', 17), ('2', 43), ('4', 1), ('3', 17), ('4', 8)]

没有使用任何导入的包

Answer 1

您可以尝试以下方法；累积值直到“键”发生变化：

list_a = [('0', 60), ('0', 5), ('2', 1), ('3', 14), ('4', 39), ('1', 17), ('2', 14), ('2', 29), ('4', 1), ('3', 1), ('3', 16), ('4', 8)]

output = []
v_accu = 0 # value accumulator
k_prev = None
for k, v in list_a:
    if k_prev is not None and k != k_prev: # can omit "is not None"
        output.append((k_prev, v_accu)) # insert the accumulated value so far
        v_accu = v # reset the accumulator
    else:
        v_accu += v # keep adding
    k_prev = k # update k_prev
output.append((k_prev, v_accu)) # final insertion
print(output)
# [('0', 65), ('2', 1), ('3', 14), ('4', 39), ('1', 17), ('2', 43), ('4', 1), ('3', 17), ('4', 8)]

但是对于一些标准模块，你可以用更简单的方式来做：

from itertools import groupby
from operator import itemgetter

list_a = [('0', 60), ('0', 5), ('2', 1), ('3', 14), ('4', 39), ('1', 17), ('2', 14), ('2', 29), ('4', 1), ('3', 1), ('3', 16), ('4', 8)]

output = [(k, sum(map(itemgetter(1), g))) for k, g in groupby(list_a, key=itemgetter(0))]
print(output)

关于 Python 元组操作

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