Python–用列表展平词典
原文:https://www . geesforgeks . org/python-flat-dictionary-with-list/
给定一个列表和字典，在列表中键的可用元素的位置展平带有键和值的字典。
输入:test _ list =[“Gfg”、“is”、“Best”、“For”、“Geeks”]，subs _ dict = {“Gfg”:7 } 输出:[‘Gfg’，7，‘is’，‘Best’，‘For’，‘Geeks’] 解释:“Gfg”被替换，后面是它在字典中的值。
输入:test _ list =[“Gfg”、“is”、“Best”、“For”、“Geeks”]，subs _ dict = {“Gfg”:7、“Best”:8 } 输出:【‘Gfg’、‘is’、‘Best’、‘For’、‘Geeks’】 解释:无替换。没有匹配的值。
方法一:使用列表理解+ get()
上述功能的组合可以用来解决这个问题。在这种情况下，我们使用 get()附加所有的键(如果存在的话)以及列表中的值。
Python 3
# Python3 code to demonstrate working of
# Flatten Dictionary with List
# Using get() + list comprehension
# initializing list
test_list = ["Gfg", "is", "Best", "For", "Geeks"]
# printing original list
print("The original list : " + str(test_list))
# initializing subs. Dictionary
subs_dict = {"Gfg" : 7, "Geeks" : 8}
# get() to perform presence checks and assign value
temp = object()
res = [ele for sub in ((ele, subs_dict.get(ele, temp))
for ele in test_list) for ele in sub if ele != temp]
# printing result
print("The list after substitution : " + str(res))
Output
The original list : ['Gfg', 'is', 'Best', 'For', 'Geeks']
The list after substitution : ['Gfg', 7, 'is', 'Best', 'For', 'Geeks', 8]
方法 2:使用 chain.from_iterable() +列表理解
这是执行这项任务的另一种方式。在这种情况下，我们形成键值对列表，如果存在就追加，如果不存在就保留元素。接下来，使用 chain.from_iterable()执行键值列表的展平。
Python 3
# Python3 code to demonstrate working of
# Flatten Dictionary with List
# Using chain.from_iterable() + list comprehension
from itertools import chain
# initializing list
test_list = ["Gfg", "is", "Best", "For", "Geeks"]
# printing original list
print("The original list : " + str(test_list))
# initializing subs. Dictionary
subs_dict = {"Gfg" : 7, "Geeks" : 8}
temp = ([ele, subs_dict[ele]] if ele in subs_dict
else [ele] for ele in test_list)
# chain.from_iterable() for flattening
res = list(chain.from_iterable(temp))
# printing result
print("The list after substitution : " + str(res))
Output
The original list : ['Gfg', 'is', 'Best', 'For', 'Geeks']
The list after substitution : ['Gfg', 7, 'is', 'Best', 'For', 'Geeks', 8]
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