注:本代码主要是为了实现多个集合之间求并集时的辅助代码,简单的举个例子来说明代码的功能。
约定:例如{11: [2, 3]}表示数据11在集合2和集合3中都存在。
现有以下数据:
d0 = {38: [2], 11: [2, 3], 22: [2, 3]}
d1 = {11: [0, 0, 1, 3], 13: [0, 0], 22: [0, 0, 3], 14: [0, 0, 1, 3]}
d2 = {32: [1, 1], 35: [1, 1], 11: [1, 1, 3], 14: [1, 1, 3], 23: [1, 1]}
d3 = {33: [3], 66: [3], 11: [3], 14: [3], 22: [3], 55: [3]}
d4 = {38: [2], 11: [2, 3], 22: [2, 3], 13: [0, 0], 14: [0, 0, 1, 3], 32: [1, 1], 35: [1, 1], 23: [1, 1], 33: [3], 66: [3], 55: [3]}
那么所有的元素所在的集合为:
38: [2], 11: [2, 3, 0, 1], 22: [2, 3, 0], 13: [0], 14: [0, 1, 3], 32: [1], 35: [1], 23: [1], 33: [3], 66: [3], 55: [3] 。
下面就是实现的代码:
def func(content):
all_dict = {}
for d in content:
for k in d.keys():
tmp_value = d[k]
if k in all_dict.keys():
tmp_values = all_dict[k]
# 更新成员数
for i in range(len(tmp_value)):
if tmp_value[i] not in tmp_values:
tmp_values.append(tmp_value[i])
all_dict[k] = tmp_values
if k not in all_dict.keys():
# 这里需要处理一种情况:如13: [0, 0],某个字典的值有重复的元素。
tmp_values = []
for i in range(len(tmp_value)):
tmp = tmp_value[i]
if tmp in tmp_values:
continue
tmp_values.append(tmp)
all_dict[k] = tmp_values
return all_dict
if __name__ == "__main__":
d0 = {38: [2], 11: [2, 3], 22: [2, 3]}
d1 = {11: [0, 0, 1, 3], 13: [0, 0], 22: [0, 0, 3], 14: [0, 0, 1, 3]}
d2 = {32: [1, 1], 35: [1, 1], 11: [1, 1, 3], 14: [1, 1, 3], 23: [1, 1]}
d3 = {33: [3], 66: [3], 11: [3], 14: [3], 22: [3], 55: [3]}
d4 = {38: [2], 11: [2, 3], 22: [2, 3], 13: [0, 0], 14: [0, 0, 1, 3], 32: [1, 1], 35: [1, 1], 23: [1, 1], 33: [3], 66: [3], 55: [3]}
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res = func([d0, d1, d2, d3, d4])
for key in res:
print (key, ':', res[key])
输出结果为:
38 : [2]
11 : [2, 3, 0, 1]
22 : [2, 3, 0]
13 : [0]
14 : [0, 1, 3]
32 : [1]
35 : [1]
23 : [1]
33 : [3]
66 : [3]
55 : [3]
标签:tmp,11,14,22,python,所在,获取,values,dict
From: https://www.cnblogs.com/Pythonmiss/p/18215840