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法学Python - 查找列表中值的索引的方法
通常，我们需要找出特定值位于其中的索引。有很多方法可以实现这一点，使用 index() 等。但是在列表中特定值有多次出现的情况下，有时需要找出所有这些值的索引。
示例
# using filter()
# initializing list
test_list = [1, 3, 4, 3, 6, 7]
# printing initial list
print ("Original list : " + str(test_list))
# using filter()
# to find indices for 3
res_list = list(filter(lambda x: test_list[x] == 3, range(len(test_list))))
# printing resultant list
print ("New indices list : " + str(res_list))
# using enumerate()
# initializing list
test_list = [1, 3, 4, 3, 6, 7]
# printing initial list
print ("Original list : " + str(test_list))
# using enumerate()
# to find indices for 3
res_list = [i for i, value in enumerate(test_list) if value == 3]
# printing resultant list
print ("New indices list : " + str(res_list))
# using list comprehension
# initializing list
test_list = [1, 3, 4, 3, 6, 7]
# printing initial list
print ("Original list : " + str(test_list))
# using list comprehension
# to find indices for 3
res_list = [i for i in range(len(test_list)) if test_list[i] == 3]
# printing resultant list
print ("New indices list : " + str(res_list))
# using naive method
# initializing list
test_list = [1, 3, 4, 3, 6, 7]
# printing initial list
print ("Original list : " + str(test_list))
# using naive method
# to find indices for 3
res_list = []
for i in range(0, len(test_list)) :
if test_list[i] == 3 :
res_list.append(i)
# printing resultant list
print ("New indices list : " + str(res_list))输出
Original list : [1, 3, 4, 3, 6, 7] New indices list : [1, 3] Original list : [1, 3, 4, 3, 6, 7] New indices list : [1, 3] Original list : [1, 3, 4, 3, 6, 7] New indices list : [1, 3] Original list : [1, 3, 4, 3, 6, 7] New indices list : [1, 3]
更新于:06-8-2020
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