python 字典的减法

da = [
    {"a": 231, "b": 456},
    {"a": 423, "b": 980},
    {"a": 846, "b": 1960},
    {}
]
lst = []
for i in range(len(da) - 1):  # 0 1
    value1 = list(da[i].values())
    value2 = list(da[i + 1].values())
    if value1 and value2:
        for j in range(len(value1)):
            lst.append(value1[j] - value2[j])
print(lst)
##########结果
[-192, -524, -423, -980]

da = [
    {"a": 231, "b": 456},
    {"a": 423, "b": 980},
    {"a": 846, "b": 1960},
    {}
]
lst = []
for i in range(len(da) - 1):  # 0 1
    value1 = list(da[i].values())
    value2 = list(da[i + 1].values())
    if value1 and value2:
        v_lst = []
        for j in range(len(value1)):
            v_lst.append(value1[j] - value2[j])
        lst.append(v_lst)
print(lst)
#############结果
[[-192, -524], [-423, -980]]

lst = []
for i in range(len(da) - 1):  # 0 1
    value1 = list(da[i].values())
    value2 = list(da[i + 1].values())
    v_lst = []
    if not value1:
        lst.append(v_lst)
    elif not value2:
        lst.append(v_lst)
    elif value1 and value2:
        for j in range(len(value1)):
            v_lst.append(value1[j] - value2[j])
        lst.append(v_lst)
print(lst)
################结果
[[], [-192, -524], [], [], []]

这种稍微麻烦一点 需要给每个字典的外层在加一层列表好用索引来

this_month_list.append(this_month_cumulative_dic)
this_month_cumulative_list.append(this_month_list)  # 各个指标的本月止累计 总和
same_month_list.append(same_month_cumulative_dic)
same_month_cumulative_list.append(same_month_list)  # 各个指标去年同月止累计 总和
for m in range(12):
    for base_dict, sub_dict in zip(this_month_cumulative_list[m], this_month_cumulative_list[m + 1]):
        this_month.append({k: v - sub_dict.get(k, 0) for k, v in base_dict.items()})

那个满足就用那个吧,能实现功能就行