创建一个迭代输入值列表的字典列表答案

【问题标题】：Create a list of Dictionaries iterating over list of input values创建一个迭代输入值列表的字典列表
【发布时间】：2021-07-20 18:43:26
【问题描述】：

我正在尝试根据单独的输入列表创建字典列表。

示例：输入

appetizer = ['soup', 'salad']
entree = ['steak', 'chicken', fish']
dessert = ['ice cream', 'flan', cake']

dict = {appetizer = [], entree = [], dessert = []}

dinner_options = []

从这里，我想创建一个包含所有可能的晚餐选项的字典列表

输出

dinner_options = [{appetizer = 'soup', entree = 'steak', dessert = 'ice cream'},
                  {appetizer = 'soup', entree = 'steak', dessert = 'flan'},
                  {appetizer = 'soup', entree = 'steak', dessert = 'cake'},
                  {appetizer = 'soup', entree = 'chicken', dessert = 'ice cream'},
                  {appetizer = 'soup', entree = 'chicken', dessert = 'flan'},
                  {appetizer = 'soup', entree = 'chicken', dessert = 'cake'},
                  {appetizer = 'soup', entree = 'fish', dessert = 'ice cream'},
                  {appetizer = 'soup', entree = 'fish', dessert = 'flan'},
                  {appetizer = 'soup', entree = 'fish', dessert = 'cake'},
                  {appetizer = 'salad', entree = 'steak', dessert = 'ice cream'},
                  {appetizer = 'salad', entree = 'steak', dessert = 'flan'},
                  {appetizer = 'salad', entree = 'steak', dessert = 'cake'},
                  {appetizer = 'salad', entree = 'chicken', dessert = 'ice cream'},
                  {appetizer = 'salad', entree = 'chicken', dessert = 'flan'},
                  {appetizer = 'salad', entree = 'chicken', dessert = 'cake'},
                  {appetizer = 'salad', entree = 'fish', dessert = 'ice cream'},
                  {appetizer = 'salad', entree = 'fish', dessert = 'flan'},
                  {appetizer = 'salad', entree = 'fish', dessert = 'cake'}]

基本上，最后我试图获得一个列表，其中包含输入中提供的列表（参数）的所有排列。

【问题讨论】：

你的输入数组总是3还是可以改变？

标签： python list dictionary while-loop nested-loops

【解决方案1】：

您只需要遍历您的列表并将可能性附加到您的 dinner_options 数组。

appetizer = ['soup', 'salad']
entree = ['steak', 'chicken', 'fish']
dessert = ['ice cream', 'flan', 'cake']

dinner_options = []

for ap in appetizer:
    for en in entree:
        for de in dessert:
            dinner_options.append({ "appetizer": ap, "entree": en, "dessert": de })

print(dinner_options)

然后它应该显示

[{'appetizer': 'soup', 'entree': 'steak', 'dessert': 'ice cream'}, {'appetizer': 'soup', 'entree': 'steak', 'dessert': 'flan'}, {'appetizer': 'soup', 'entree': 'steak', 'dessert': 'cake'}, {'appetizer': 'soup', 'entree': 'chicken', 'dessert': 'ice cream'}, {'appetizer': 'soup', 'entree': 'chicken', 'dessert': 'flan'}, {'appetizer': 'soup', 'entree': 'chicken', 'dessert': 'cake'}, {'appetizer': 'soup', 'entree': 'fish', 'dessert': 'ice cream'}, {'appetizer': 'soup', 'entree': 'fish', 'dessert': 'flan'}, {'appetizer': 'soup', 'entree': 'fish', 'dessert': 'cake'}, {'appetizer': 'salad', 'entree': 'steak', 'dessert': 'ice cream'}, {'appetizer': 'salad', 'entree': 'steak', 'dessert': 'flan'}, {'appetizer': 'salad', 'entree': 'steak', 'dessert': 'cake'}, {'appetizer': 'salad', 'entree': 'chicken', 'dessert': 'ice cream'}, {'appetizer': 'salad', 'entree': 'chicken', 'dessert': 'flan'}, {'appetizer': 'salad', 'entree': 'chicken', 'dessert': 'cake'}, {'appetizer': 'salad', 'entree': 'fish', 'dessert': 'ice cream'}, {'appetizer': 'salad', 'entree': 'fish', 'dessert': 'flan'}, {'appetizer': 'salad', 'entree': 'fish', 'dessert': 'cake'}]

【讨论】：

【解决方案2】：

你可以使用itertools.product:

from itertools import product

appetizer = ['soup', 'salad']
entree = ['steak', 'chicken', 'fish']
dessert = ['ice cream', 'flan', 'cake']

# Here we create an iterable from the objects we want to obtain
# different combinations.
to_combine = (appetizer, entree, dessert)

# Next, we need to define the order in which those appear
# as those will be used as the keys in the individual
# dictionary objects.
order = 'appetizer,entree,dessert'.split(',')
dinner_options = list()

# Now we loop over the unpacked tuple and build each individual
# dictionary.
for items in product(*to_combine):
    # This creates the individual objects which are of the form:
    # `{'appetizer': '<a>', 'entree': '<e>', 'dessert': '<d>'}
    d = {key:value for key, value in zip(order, items)}
    dinner_options.append(d)

这会产生所需的输出。

【讨论】：

【解决方案3】：

我认为这是最明显也是最耗时的答案

appetizer = ['soup', 'salad']
entree = ['steak', 'chicken', 'fish']
dessert = ['ice cream', 'flan', 'cake']

# dict = {appetizer = [], entree = [], dessert = []}

dinner_options = []

for start in appetizer:
    for mid in entree:
        for end in dessert:
            data = dict(appetizer=start, entree=mid, dessert=end)
            dinner_options.append(data)

print(dinner_options)

【讨论】：