Question

我想我有两个问题。首先是“我走的路是否正确，如果不正确，我应该怎么做？”第二个是“如何停止覆盖值”。

在下面的代码中，如何使“ monthdict”的输出与“ report”相同？

报告是一个值列表，其中（例如）1月1日星期一的面包是1，而2月1日星期一的面包是31。当我尝试以编程方式重新创建时，它使所有面包都变成53。

我在现实世界中的使用情况是从某个API传入的实际数据，一次只能访问7个项目。因此，我需要依次调用1月1日星期一，1月1日星期二，1月星期三星期三，1月2日星期一星期一等，并返回值

report = {
    "january" : {
        "week1" : {
            "monday" : {
                 "bread" : 1,
                 "cheese" : 2,
                 "milk" : 3
                 },
            "tueday" : {
                 "bread" : 4,
                 "cheese" : 5,
                 "milk" : 6
                 },
            "wednesday" : {
                 "bread" : 7,
                 "cheese" : 8,
                 "milk" : 9
                 },
        },
        "week2" : { 
            "monday" : {
                 "bread" : 11,
                 "cheese" : 12,
                 "milk" : 13
                 },
            "tueday" : {
                 "bread" : 14,
                 "cheese" : 15,
                 "milk" : 16
                 },
            "wednesday" : {
                 "bread" : 17,
                 "cheese" : 18,
                 "milk" : 19
                 },
        },
        "week3" : { 
            "monday" : {
                 "bread" : 21,
                 "cheese" : 22,
                 "milk" : 23
                 },
            "tueday" : {
                 "bread" : 24,
                 "cheese" : 25,
                 "milk" : 26
                 },
            "wednesday" : {
                 "bread" : 27,
                 "cheese" : 28,
                 "milk" : 29
                 },
        },
    },
    "february" : {
        "week1" : { 
            "monday" : {
                 "bread" : 31,
                 "cheese" : 32,
                 "milk" : 33
                 },
            "tueday" : {
                 "bread" : 34,
                 "cheese" : 35,
                 "milk" : 36
                 },
            "wednesday" : {
                 "bread" : 37,
                 "cheese" : 38,
                 "milk" : 39
                 },
        },
        "week2" : { 
            "monday" : {
                 "bread" : 111,
                 "cheese" : 112,
                 "milk" : 113
                 },
            "tueday" : {
                 "bread" : 114,
                 "cheese" : 115,
                 "milk" : 116
                 },
            "wednesday" : {
                 "bread" : 117,
                 "cheese" : 118,
                 "milk" : 119
                 },
        },
        "week3" : { 
            "monday" : {
                 "bread" : 121,
                 "cheese" : 122,
                 "milk" : 123
                 },
            "tueday" : {
                 "bread" : 124,
                 "cheese" : 125,
                 "milk" : 126
                 },
            "wednesday" : {
                 "bread" : 127,
                 "cheese" : 128,
                 "milk" : 129
                 },
        }
    }
}

print("\nreport:\n", report, "\n\n")

months = ["january", "february"]
weeks = ["week1", "week2", "week3"]
days = ["monday", "tuesday", "wednesday"]
food = ["bread", "cheese", "milk"]
values = []

for i in range(1,129):
    values.append(i)

x = 0

fooddict = {}
daydict = {}
weekdict = {}
monthdict= {}

for m in months:
    for w in weeks:
        for d in days:
            for f in food:
                fooddict[f] = values[x]
                x += 1
                print(x, ") adding ", values[x], " to ", f)
            daydict[d] = fooddict
        weekdict[w] = daydict
    monthdict[m] = weekdict

print(monthdict)

Answer 1

尝试

monthdict= {}
for m in months:
    weekdict = {}
    for w in weeks:
        daydict = {}
        for d in days:
            fooddict = {}
            for f in food:
                fooddict[f] = values[x]
                x += 1
                print(x, ") adding ", values[x], " to ", f)
            daydict[d] = fooddict
        weekdict[w] = daydict
    monthdict[m] = weekdict

否则，您将共享前一个月，一周和几天中的值（相反，对于每个新迭代，它们应以空值开头）。

蟒蛇。无法通过for循环将值添加到嵌套词典

1 个答案: