有没有办法返回在 any() 函数中返回 True 的元素的索引？

Question

（这只是我想解决的一个例子）

array = ["hello", "hi"]

statement = input()

condition = any(statement in elm for elm in array)

有没有办法返回返回 True 的元素的索引，或者我应该只使用 for 循环？

Answer 1

其实你想要array中声明的index

array = ["hello", "hi"]
statement = input("Give a word: ")
condition = array.index(statement) if statement in array else -1
print(condition)

演示

Give a word: hello
0
Give a word: hi
1
Give a word: Hi
-1

Answer 2

下次搜索函数属性。我基于Python documents编写示例代码

array = ["hello", "hi", "example", "to get", "index from array"]

def SearchIndex():
    statement = input()
    #// Check if input is item on list
    if statement in array: print(">>", array.index(statement))

    #// Other search if input is part of item in list
    else:
        part_of = False
        for item in array:
            if statement in item:
                print('>> {0} is part of "{2}" -> {1}'.format(statement, array.index(item), array[array.index(item)]))
                #// If founded signal
                part_of = True
        #// If nout found (signal is false)
        if part_of == False: print(f">> {statement} is not on list...")

while 1:
    SearchIndex()