如何找到字符串中任何一组字符的第一个索引 [英] How to find the first index of any of a set of characters in a string

问题描述

我想找到字符串中任何特殊"字符第一次出现的索引，如下所示:

<预><代码>>>>"Hello world!".index([' ', '!'])5

...除了这不是有效的 Python 语法.当然，我可以写一个函数来模拟这种行为:

def first_index(s, characters):我 = []对于 c 字符:尝试:i.append(s.index(c))除了值错误:经过如果不是我:引发值错误返回 min(i)

我也可以使用正则表达式，但两种解决方案似乎都有些矫枉过正.在 Python 中是否有任何理智"的方法可以做到这一点?

解决方案

您可以使用 enumerate 和 next 和 生成器表达式，如果s中没有字符出现则获取第一个匹配或返回None:

s = "Hello world!"st = {"!"," "}ind = next((i for i, ch in enumerate(s) if ch in st),None)打印(索引)

如果没有匹配项，您可以将接下来要传递的任何值作为默认返回值传递.

如果你想使用一个函数并引发一个 ValueError:

def first_index(s, characters):st = 设置(字符)ind = next((i for i, ch in enumerate(s) if ch in st), None)如果 ind 不是 None:返回工业引发值错误

对于较小的输入，使用集合不会有太大区别，但对于大字符串，它会更有效.

一些时间:

在字符串中，字符集的最后一个字符:

In [40]: s = "Hello world!"* 100在 [41] 中:字符串 = s在 [42] 中:%%timeitst = {"x","y","!"}next((i for i, ch in enumerate(s) if ch in st), None)....:1000000 个循环，最好的 3 个:每个循环 1.71 µs在 [43] 中:%%timeit特价 = ['x', 'y', '!']min(map(lambda x: (string.index(x) if (x in string) else len(string)), specials))....:100000 个循环，最好的 3 个:每个循环 2.64 µs

不在字符串中，更大的字符集:

在 [44]: %%timeitst = {"u","v","w","x","y","z"}next((i for i, ch in enumerate(s) if ch in st), None)....:1000000 个循环，最好的 3 个:每个循环 1.49 µs在 [45] 中:%%timeitspecials = ["u","v","w","x","y","z"]min(map(lambda x: (string.index(x) if (x in string) else len(string)), specials))....:100000 个循环，最好的 3 个:每个循环 5.48 µs

在字符串中字符集的第一个字符:

在 [47]: %%timeit特价 = ['H', 'y', '!']min(map(lambda x: (string.index(x) if (x in string) else len(string)), specials))....:100000 个循环，最好的 3 个:每个循环 2.02 µs在 [48] 中:%%timeitst = {"H","y","!"}next((i for i, ch in enumerate(s) if ch in st), None)....:1000000 个循环，最好的 3 个:每个循环 903 ns

I'd like to find the index of the first occurrence of any "special" character in a string, like so:

>>> "Hello world!".index([' ', '!'])
5

…except that's not valid Python syntax. Of course, I can write a function that emulates this behavior:

def first_index(s, characters):
    i = []
    for c in characters:
        try:
            i.append(s.index(c))
        except ValueError:
            pass
    if not i:
        raise ValueError
    return min(i)

I could also use regular expressions, but both solutions seem to be a bit overkill. Is there any "sane" way to do this in Python?

解决方案

You can use enumerate and next with a generator expression, getting the first match or returning None if no character appears in s:

s = "Hello world!"

st = {"!"," "}
ind = next((i for i, ch  in enumerate(s) if ch in st),None)
print(ind)

You can pass any value you want to next as a default return value if there is no match.

If you want to use a function and raise a ValueError:

def first_index(s, characters):
    st = set(characters)
    ind = next((i for i, ch in enumerate(s) if ch in st), None)
    if ind is not None:
        return ind
    raise ValueError

For smaller inputs using a set won't make much if any difference but for large strings it will be a more efficient.

Some timings:

In the string, last character of character set:

In [40]: s = "Hello world!" * 100    
In [41]: string = s    
In [42]: %%timeit
st = {"x","y","!"}
next((i for i, ch in enumerate(s) if ch in st), None)
   ....: 
1000000 loops, best of 3: 1.71 µs per loop    
In [43]: %%timeit
specials = ['x', 'y', '!']
min(map(lambda x: (string.index(x) if (x in string) else len(string)), specials))
   ....: 
100000 loops, best of 3: 2.64 µs per loop

Not in the string, larger character set:

In [44]: %%timeit
st = {"u","v","w","x","y","z"}
next((i for i, ch in enumerate(s) if ch in st), None)
   ....: 
1000000 loops, best of 3: 1.49 µs per loop

In [45]: %%timeit
specials = ["u","v","w","x","y","z"]
min(map(lambda x: (string.index(x) if (x in string) else len(string)), specials))
   ....: 
100000 loops, best of 3: 5.48 µs per loop

In string an very first character of character set:

In [47]: %%timeit
specials = ['H', 'y', '!']
min(map(lambda x: (string.index(x) if (x in string) else len(string)), specials))
   ....: 
100000 loops, best of 3: 2.02 µs per loop

In [48]: %%timeit
st = {"H","y","!"}
next((i for i, ch in enumerate(s) if ch in st), None)
   ....: 
1000000 loops, best of 3: 903 ns per loop

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