[英]Django, how to sum queryset value in dictionary values
我有以下查詢集字典:
{'Key_1': [100.0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
'Key_2': [103.0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]}
其中我將我的產品類別作為鍵,並將項目 12 值作為每個月的總和。 我想計算每個項目的所有鍵的交叉和,如下例所示:
{'Total': [203.0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]}
我怎么能得到它?
您可以使用zip
、 sum
和列表理解:
d = {
'Key_1': [100.0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
'Key_2': [103.0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
}
sum_dict = {
'Total': [sum(t) for t in zip(*d.values())],
}
# sum_dict = {'Total': [203.0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]}
[sum(t) for t in zip(*d.values())]
的逐步說明:
[sum(t) for t in zip([100.0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [103.0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])]
[sum(t) for t in [(100.0, 103.0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0), (0, 0)]]
[sum((100.0, 103.0)), sum((0, 0)), sum((0, 0)), sum((0, 0)), sum((0, 0)), sum((0, 0)), sum((0, 0)), sum((0, 0)), sum((0, 0)), sum((0, 0)), sum((0, 0)), sum((0, 0))]
[203.0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
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