python - 为什么Python(3.7)不断覆盖 key :value pairs in my nested dictionary?

Question

我想以此作为序言，说我通过编写以下内容在 python 终端中复制了代码的精神:

A = {}
A["Q"] = {}
A["Q"]["creation"] = {}
A["Q"]["creation"]["reactants_1"] = ["R1","R2"]

打印 A 给出了人们所期望的结果:

{"Q" : {"creation" : {"reactants_1" : ["R1","R2"]}}}

然后添加:

A["Q"]["creation"]["reactants_2"] = ["R3","R4"]

产量:

{"Q" : {"creation" : {"reactants_1" : ["R1","R2"], "reactants_2" : ["R3","R4"]}}}

但是，我在下面显示了一个函数，当正确运行时，它会分配前几个键:值对，但是当它循环到 x 的第二个值时，它会用新的键:值对替换之前写入的第一个键:值对关键是不同的。

使用上面的例子我们只会得到:

{"Q" : {"creation" : {"reactants_2" : ["R3","R4"]}}}

Reactions 是一个包含“e+e+p=e+H_1”之类的数组，格式如下:

["e+e+p=e+H_1", "e+e+p=e+H_2",...,"e+H_10=e+e+p"]

Species 是一个数组，例如:

[["e", 100000],["p", 100000],...]

现在我们只关心字符串部分。

diff_input 是一个空字典

reactions_const 包含，对于每个 react ，左侧和右侧分开 - 在函数早期被视为 [x][0][0] 和 [x][0][1] 以及一些其他信息目前并不重要。

rates_names 是每个 react 的唯一标识符数组，我稍后可以使用，因此使用字典方法。

打印语句都是我试图找出它不起作用的原因

def rate_eqns(diff_input, reactions, species, reactions_const, 
          rates_names):

for x in range(len(reactions)):
    # for each reaction
    # split the left and right hand side up into lists of their 
    # respective species involved for counting later

    print("reaction: " + str(x) + " " + str(reactions[x]))

    species_lhs = reactions_const[x][0][0].split('+')
    print("LHS = " + str(species_lhs))

    species_rhs = reactions_const[x][0][1].split('+')

    for y in range(len(species)):  
        # For each species, create a sub-dictionary
        diff_input[species[y][0]] = {}

        # create sub-dictionaries in each species for creation, destruction and preservation/neutral paths
        diff_input[species[y][0]]["destruction"] = {}
        diff_input[species[y][0]]["creation"] = {}
        diff_input[species[y][0]]["balanced"] = {}

        # check if species occurs in each reaction

        if species[y][0] in reactions[x][0]:

            # if you start with more of it than you finish its destruction
            if species_lhs.count(species[y][0]) > species_rhs.count(species[y][0]):

                # if true: add an entry to the dictionary which holds the reaction identifier
                # bound to the destruction/creation/balanced identifier bound to the species identifier.
                print("species:" + str(species[y][0]) + " found net loss from reactants")
                print("LHS = " + str(species_lhs))
                print("RHS = " + str(species_rhs))
                print("reaction designation = " + str(rates_names[x]) + " Destruction of species")
                print("-------------")
                diff_input[species[y][0]]["destruction"][rates_names[x]] = species_lhs

            elif species_lhs.count(species[y][0]) == species_rhs.count(species[y][0]):
                print("species:" + str(species[y][0]) + " found no change in number")
                print("LHS = " + str(species_lhs))
                print("RHS = " + str(species_rhs))
                print("reaction designation = " + str(rates_names[x]) + " preservation of species")
                diff_input[species[y][0]]["balanced"][rates_names[x]] = species_lhs

            elif species_lhs.count(species[y][0]) < species_rhs.count(species[y][0]):
                print("species:" + str(species[y][0]) + " found net gain from reactants")
                print("LHS = " + str(species_lhs))
                print("RHS = " + str(species_rhs))
                print("reaction designation = " + str(rates_names[x]) + " creation of species")
                diff_input[species[y][0]]["creation"][rates_names[x]] = species_lhs

        # else:
            # print(str(species[y][0]) + " not found in reaction")
        print(diff_input)
        a = input("press return to continue")
with open('diff_input.txt', 'w') as file:
    file.write(str(diff_input))

return diff_input

文件保存部分是可选的，是否有其他人遇到过用新键覆盖现有键的字典？

感谢您的耐心等待，并且感谢您对我的格式设置提出的任何建议(我试图使其尽可能好，而不包括脚本的其余部分)

Answer 1

species[1][0] 的实际值必须恰好等于 species[0][0]，这样当它循环到x 的第二个值，赋值 diff_input[species[y][0]] = {} 将覆盖上一次迭代的子字典，因为 species [y][0] 保持不变。

更简单地说，使用示例代码，在外循环中进行以下初始化:

A["Q"] = {}
A["Q"]["creation"] = {}

所以即使你的内部循环为子字典分配了一些值:

A["Q"]["creation"]["reactants_1"] = ["R1","R2"]

A["Q"] = {} 只要 "Q" 继续作为作业的主键，就会在下一次迭代中覆盖它。