python - python3 - learning about searching, this very simple example does not work right

Question

here's the code:

def findsort(something, alist):
    for item in alist:
        if item == something:
            return "found"
        else:
            return "not found"

def main():

    print(findsort(6, [3, 6, 1, 8, 11, 14]))

main()

for some reason, this doesn't work the way i thought it should work.

when i run this, it will say "not found." however, if i change the value i want to find to the first item in the list, the 3, it comes back as "found."

i have tried this with strings, and i get the same results.

can someone please tell me what i am doing wrong?

Answer 1

因为如果在第一次迭代中项目不匹配，则进入 else 分支返回“未找到”，从而退出循环。

尝试这个：

def findsort(something, alist):
    for item in alist:
        if item == something:
            return "found"
    return "not found"

或者简单地说：

def findsort(something, alist):
    return "found" if something in alist else "not found"

Answer 2

@hyperboreus 指出了错误的原因（在else看到所有项目之前执行的分支）。

要在排序（“有序”）列表中查找项目，您可以使用执行二进制搜索（）而不是线性搜索（ ）的bisect模块，例如，对于一百万个项目，二进制搜索将需要大约几十个操作，而对于线性搜索。O(log(n)item in alistO(n)

from bisect import bisect

findsort = lambda x, L: "found" if L and L[bisect(L,x) - 1] == x else "not found"