Python – 按键的第 i 个索引值对字典列表进行排序
给定字典列表，根据 Key 的第 i 个索引值对字典进行排序
Input : [{“Gfg” : “Best”, “for” : “Geeks”}, {“Gfg” : “Good”, “for” : “Me”}, {“Gfg” : “Better”, “for” : “All”}], K = “Gfg”, i = 1
Output : [{‘Gfg’: ‘Best’, ‘for’: ‘Geeks’}, {‘Gfg’: ‘Better’, ‘for’: ‘All’}, {‘Gfg’: ‘Good’, ‘for’: ‘Me’}]
Explanation : Sort in order of e = e < o, as 1st index element of "Gfg"'s value.
Input : [{“Gfg” : “Best”, “for” : “Geeks”}, {“Gfg” : “Good”, “for” : “Me”}, {“Gfg” : “Better”, “for” : “All”}], K = “Gfg”, i = 0
Output : [{‘Gfg’: ‘Best’, ‘for’: ‘Geeks’}, {‘Gfg’: ‘Better’, ‘for’: ‘All’}, {‘Gfg’: ‘Good’, ‘for’: ‘Me’}]
Explanation : Sort in order of B = B < G, as 1st index element of "Gfg"'s value.
方法 #1：使用 sort() + lambda
上述功能的组合可以用来解决这个问题。在此，我们在“key”参数驱动条件下使用 sort() 和 lambda函数执行排序任务。
Python3
# Python3 code to demonstrate working of
# Sort Dictionary List by Key's ith Index value
# Using sort() + lambda
# initializing lists
test_list = [{"Gfg" : "Best", "for" : "Geeks"},
{"Gfg" : "Good", "for" : "Me"},
{"Gfg" : "Better", "for" : "All"}]
# printing original list
print("The original list : " + str(test_list))
# initializing K
K = "Gfg"
# initializing i
i = 2
# using sort to perform sort(), lambda
# function drives conditions
res = sorted(test_list, key = lambda sub: sub[K][i])
# printing result
print("List after sorting : " + str(res))Python3
# Python3 code to demonstrate working of
# Sort Dictionary List by Key's ith Index value
# Using sort() + lambda + get()
# initializing lists
test_list = [{"Gfg" : "Best", "for" : "Geeks"},
{"Gfg" : "Good", "for" : "Me"},
{"Gfg" : "Better", "for" : "All"}]
# printing original list
print("The original list : " + str(test_list))
# initializing K
K = "Gfg"
# initializing i
i = 2
# using sort to perform sort(), lambda
# function drives conditions, get() used to
# avoid missing key error
res = sorted(test_list, key = lambda sub: sub.get(K)[i])
# printing result
print("List after sorting : " + str(res))The original list : [{‘Gfg’: ‘Best’, ‘for’: ‘Geeks’}, {‘Gfg’: ‘Good’, ‘for’: ‘Me’}, {‘Gfg’: ‘Better’, ‘for’: ‘All’}]
List after sorting : [{‘Gfg’: ‘Good’, ‘for’: ‘Me’}, {‘Gfg’: ‘Best’, ‘for’: ‘Geeks’}, {‘Gfg’: ‘Better’, ‘for’: ‘All’}]
方法 #2：使用 sort() + lambda + get()
以上功能的组合也可以解决这个问题。这只是上述方法的轻微变化。在此，我们使用 get() 来避免特定记录中不存在密钥的机会。
Python3
# Python3 code to demonstrate working of
# Sort Dictionary List by Key's ith Index value
# Using sort() + lambda + get()
# initializing lists
test_list = [{"Gfg" : "Best", "for" : "Geeks"},
{"Gfg" : "Good", "for" : "Me"},
{"Gfg" : "Better", "for" : "All"}]
# printing original list
print("The original list : " + str(test_list))
# initializing K
K = "Gfg"
# initializing i
i = 2
# using sort to perform sort(), lambda
# function drives conditions, get() used to
# avoid missing key error
res = sorted(test_list, key = lambda sub: sub.get(K)[i])
# printing result
print("List after sorting : " + str(res))
The original list : [{‘Gfg’: ‘Best’, ‘for’: ‘Geeks’}, {‘Gfg’: ‘Good’, ‘for’: ‘Me’}, {‘Gfg’: ‘Better’, ‘for’: ‘All’}]
List after sorting : [{‘Gfg’: ‘Good’, ‘for’: ‘Me’}, {‘Gfg’: ‘Best’, ‘for’: ‘Geeks’}, {‘Gfg’: ‘Better’, ‘for’: ‘All’}]