Question

因此，如果我想找到相同的键并将这些值加到字典列表中。

如果我找到键名“ hi”，

[{'hello':10,'hi':2},{'hi':3}]将返回5

Answer 1

sum和列表理解可以很容易地解决这个问题。

x = [{'hello': 10, 'hi': 2}, {'hi': 3}, {'nohihere': 8}]

all_hi = sum(d.get('hi', 0) for d in x)

print(all_hi)

Answer 2

您可以尝试以下方法：

dlist = [{'hello':10,'hi':2},{'hi':3}]
sum(map(lambda x: x.get('hi',0), dlist))

Answer 3

您只需要检查每个对象是否具有带有“ hi”的键

list = [{'hello':10,'hi':2},{'hi':3},{'nothi':4}]

sum = 0

for i in list:
  if 'hi' in i:
     sum += i['hi']

print(sum)

Answer 4

这是另一种解决方案

my_dict = [{'hello':10,'hi':2},{'hi':3}]
common_keys = set.intersection(*map(set, my_dict))
summed_dict = {key: sum(d[key] for d in my_dict) for key in common_keys}
print(summed_dict)

o / p-{'hi': 5}

否则您可以使用此

import functools as ft
my_dict = [{'hello':10,'hi':2},{'hi':3}]
print({k: sum(d[k] for d in my_dict) for k in ft.reduce(set.intersection, map(set, my_dict))})

o / p-{'hi': 5}

相同的关键发现值

4 个答案: