创建 2 个列表的 Pythonic 方法，其中一个是一组重复的字符串并变成字典

Question

我有一个 'widget types'、["C"、"A"、"B"] 的列表，需要为这些类型制作一个字典以对应于它们各自的 ID，1='C', 2='A', 3='B', 4='C', 5='A', 6='B', 7='C' ， ETC。我已经知道该怎么做，我只是想知道是否有比下面更优雅的 pythonic 方式来实现它，而且 dict/lists 的长度可扩展也很好，即在示例中它是10，但实际上它会有数百个小部件。

现有工作代码

wid_letter = ['C', 'A', 'B']
widgets_per_row = 10
wid_letter = wid_letter * widgets_per_row 
#print("wid_letter : ", wid_letter, '\n')



ID = (list(range(1,widgets_per_row  + 1,1)))
wid_type = []

# Make list of widget_letters to match length of widget ID
for i in ID:
    wid_type.append(wid_letter[i-1])
print(len(wid_type))

# Turn the two lists into a dict
wid_ID_dict = dict(zip(ID, wid_type))
print('\n'*2, wid_ID_dict,'\n')

print("Widget_type: ", wid_ID_dict[1])

Answer 1

试试：

wid_ID_dict = dict(enumerate(wid_letter, start=1))

Answer 2

这会导致相同的结果，这是您要找的吗？

from itertools import cycle

wid_letter = ['C', 'A', 'B']
letter_cycle = cycle(wid_letter)
widgets_per_row = 10

wid_ID_dict = {i: next(letter_cycle) for i in range(1,widgets_per_row+1)}

print(wid_ID_dict)

Answer 3

您可以只使用 module 来获取所需字母的索引，而不是覆盖您的列表。然后简单的字典理解就可以完成工作：

wid_letter = ['C', 'A', 'B']
widgets_per_row = 10

wid_ID_dict = {i + 1: wid_letter[i%len(wid_letter)] for i in range(widgets_per_row)}

结果如下所示：

{1: 'C', 2: 'A', 3: 'B', 4: 'C', 5: 'A', 6: 'B', 7: 'C', 8: 'A', 9: 'B', 10: 'C'}

Answer 4

wid = {i:w for i,w in enumerate(wid_letter[:widgets_per_row],1)}

上面的字典理解可能是你的任务，wid在你的代码中是wid_ID_dict。

创建 2 个列表的 Pythonic 方法，其中一个是一组重复的字符串并变成字典

Pythonic method to create 2 lists, one of which is a repeating set of strings and turn into a dict

python

zip

dictionary

list