每第 n 个字符拆分字符串

2024-11-25 08:50:00
admin
原创
163
摘要:问题描述:如何按第 n 个字符分割一个字符串?'1234567890' → ['12', '34', '56', '78', '90'] 对于带有列表的相同问题,请参阅如何将列表拆分成大小相同的块?。解决方案 1:>>> line = '1234567890' >>>...

问题描述:

如何按第 n 个字符分割一个字符串?

'1234567890'   →   ['12', '34', '56', '78', '90']

对于带有列表的相同问题,请参阅如何将列表拆分成大小相同的块?。


解决方案 1:

>>> line = '1234567890'
>>> n = 2
>>> [line[i:i+n] for i in range(0, len(line), n)]
['12', '34', '56', '78', '90']

解决方案 2:

为了完整起见,您可以使用正则表达式来执行此操作:

>>> import re
>>> re.findall('..','1234567890')
['12', '34', '56', '78', '90']

对于奇数个字符,您可以这样做:

>>> import re
>>> re.findall('..?', '123456789')
['12', '34', '56', '78', '9']

您还可以执行以下操作,以简化更长块的正则表达式:

>>> import re
>>> re.findall('.{1,2}', '123456789')
['12', '34', '56', '78', '9']

re.finditer如果字符串很长,您可以使用它来逐块生成。

解决方案 3:

Python 中已经有一个内置函数用于实现这一点。

>>> from textwrap import wrap
>>> s = '1234567890'
>>> wrap(s, 2)
['12', '34', '56', '78', '90']

这是文档字符串wrap所说的内容:

>>> help(wrap)
'''
Help on function wrap in module textwrap:

wrap(text, width=70, **kwargs)
    Wrap a single paragraph of text, returning a list of wrapped lines.

    Reformat the single paragraph in 'text' so it fits in lines of no
    more than 'width' columns, and return a list of wrapped lines.  By
    default, tabs in 'text' are expanded with string.expandtabs(), and
    all other whitespace characters (including newline) are converted to
    space.  See TextWrapper class for available keyword args to customize
    wrapping behaviour.
'''

解决方案 4:

将元素分组为 n 长度组的另一种常见方法:

>>> s = '1234567890'
>>> map(''.join, zip(*[iter(s)]*2))
['12', '34', '56', '78', '90']

此方法直接来自于 的文档zip()

解决方案 5:

我认为这比 itertools 版本更短且更易读:

def split_by_n(seq, n):
    '''A generator to divide a sequence into chunks of n units.'''
    while seq:
        yield seq[:n]
        seq = seq[n:]

print(list(split_by_n('1234567890', 2)))

解决方案 6:

使用PyPI 中的more-itertools:

>>> from more_itertools import sliced
>>> list(sliced('1234567890', 2))
['12', '34', '56', '78', '90']

解决方案 7:

我喜欢这个解决方案:

s = '1234567890'
o = []
while s:
    o.append(s[:2])
    s = s[2:]

解决方案 8:

您可以使用grouper()以下菜谱itertools

Python 2.x:

from itertools import izip_longest    

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

Python 3.x:

from itertools import zip_longest

def grouper(iterable, n, *, incomplete='fill', fillvalue=None):
    "Collect data into non-overlapping fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, fillvalue='x') --> ABC DEF Gxx
    # grouper('ABCDEFG', 3, incomplete='strict') --> ABC DEF ValueError
    # grouper('ABCDEFG', 3, incomplete='ignore') --> ABC DEF
    args = [iter(iterable)] * n
    if incomplete == 'fill':
        return zip_longest(*args, fillvalue=fillvalue)
    if incomplete == 'strict':
        return zip(*args, strict=True)
    if incomplete == 'ignore':
        return zip(*args)
    else:
        raise ValueError('Expected fill, strict, or ignore')

这些函数内存效率高,并且适用于任何可迭代对象。

解决方案 9:

这可以通过一个简单的 for 循环来实现。

a = '1234567890a'
result = []

for i in range(0, len(a), 2):
    result.append(a[i : i + 2])
print(result)

输出看起来像 ['12', '34', '56', '78', '90', 'a']

解决方案 10:

我陷入了同样的境地。

这对我有用:

x = "1234567890"
n = 2
my_list = []
for i in range(0, len(x), n):
    my_list.append(x[i:i+n])
print(my_list)

输出:

['12', '34', '56', '78', '90']

解决方案 11:

尝试一下:

s = '1234567890'
print([s[idx:idx+2] for idx in range(len(s)) if idx % 2 == 0])

输出:

['12', '34', '56', '78', '90']

解决方案 12:

尝试以下代码:

from itertools import islice

def split_every(n, iterable):
    i = iter(iterable)
    piece = list(islice(i, n))
    while piece:
        yield piece
        piece = list(islice(i, n))

s = '1234567890'
print list(split_every(2, list(s)))

解决方案 13:

与往常一样,对于那些喜欢单行的人来说:

n = 2  
line = "this is a line split into n characters"  
line = [line[i * n:i * n+n] for i, blah in enumerate(line[::n])]

解决方案 14:

>>> from functools import reduce
>>> from operator import add
>>> from itertools import izip
>>> x = iter('1234567890')
>>> [reduce(add, tup) for tup in izip(x, x)]
['12', '34', '56', '78', '90']
>>> x = iter('1234567890')
>>> [reduce(add, tup) for tup in izip(x, x, x)]
['123', '456', '789']

解决方案 15:

短字符串的简单递归解决方案:

def split(s, n):
    if len(s) < n:
        return []
    else:
        return [s[:n]] + split(s[n:], n)

print(split('1234567890', 2))

或者采用以下形式:

def split(s, n):
    if len(s) < n:
        return []
    elif len(s) == n:
        return [s]
    else:
        return split(s[:n], n) + split(s[n:], n)

,更明确地说明了递归方法中典型的分而治之模式(尽管实际上没有必要这样做)

解决方案 16:

more_itertools.sliced之前已经提到过。下面是库中的另外四个选项more_itertools

s = "1234567890"

["".join(c) for c in mit.grouper(2, s)]

["".join(c) for c in mit.chunked(s, 2)]

["".join(c) for c in mit.windowed(s, 2, step=2)]

["".join(c) for c in  mit.split_after(s, lambda x: int(x) % 2 == 0)]

后面每个选项都会产生以下输出:

['12', '34', '56', '78', '90']

讨论选项的文档:grouper,,,chunked`windowed`split_after

解决方案 17:

解决方案如下groupby

from itertools import groupby, chain, repeat, cycle

text = "wwworldggggreattecchemggpwwwzaz"
n = 3
c = cycle(chain(repeat(0, n), repeat(1, n)))
res = ["".join(g) for _, g in groupby(text, lambda x: next(c))]
print(res)

输出:

['www', 'orl', 'dgg', 'ggr', 'eat', 'tec', 'che', 'mgg', 'pww', 'wza', 'z']

解决方案 18:

另一种解决方案是使用groupbyindex//n作为对字母进行分组的键:

from itertools import groupby

text = "abcdefghij"
n = 3

result = []
for idx, chunk in groupby(text, key=lambda x: x.index//n):
    result.append("".join(chunk))

# result = ['abc', 'def', 'ghi', 'j']

解决方案 19:

从 Python 3.12 开始,该itertools库现在包含迭代器batched()

>>> from itertools import batched
>>> s = '1234567890'
>>> [''.join(batch) for batch in batched(s, 2)]
['12', '34', '56', '78', '90']

解决方案 20:

这些答案都很好,而且可行,但是语法太神秘了...为什么不写一个简单的函数呢?

def SplitEvery(string, length):
    if len(string) <= length: return [string]        
    sections = len(string) / length
    lines = []
    start = 0;
    for i in range(sections):
        line = string[start:start+length]
        lines.append(line)
        start += length
    return lines

简单来说就是:

text = '1234567890'
lines = SplitEvery(text, 2)
print(lines)

# output: ['12', '34', '56', '78', '90']

解决方案 21:

您可以在 Github 上找到包含更新解决方案的完整文章。

注意:解决方案是针对 Python3.10+ 编写的

使用列表推导和切片:这是一种简单直接的方法,我们可以使用 Python 的切片功能将字符串拆分为 n 个字符的块。我们可以使用列表推导以步长 n 迭代字符串,并将字符串从当前索引切片到当前索引加 n。

def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters.

    This function uses list comprehension and slicing to split the string into groups.
    It includes error handling to check if `n` is a positive integer.

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        list[str]: A list of strings, where each string is a group of `n` consecutive characters from the input string.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Use list comprehension and slicing to split the string into groups of `n` characters.
    return [s[i:i + n] for i in range(0, len(s), n)]

使用 re (regex) 模块:Python 的 re 模块提供了一个名为 findall() 的函数,可用于查找字符串中某个模式的所有出现位置。我们可以将此函数与匹配任意 n 个字符的正则表达式一起使用,以将字符串拆分为 n 个字符的块。

import re

def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters.

    This function uses the `re.findall()` function from the `re` (regex) module to solve the problem.
    It includes error handling to check if `n` is a positive integer.

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        list[str]: A list of strings, where each string is a group of `n` consecutive characters from the input string.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Use `re.findall()` to split the string into groups of `n` characters.
    return re.findall(f'.{{1,{n}}}', s)

使用 textwrap 模块:Python 中的 textwrap 模块提供了一个名为 wrap() 的函数,该函数可用于将字符串拆分为指定宽度的输出行列表。我们可以使用此函数将字符串拆分为 n 个字符的块。

import textwrap

def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters.

    This function uses the `textwrap.wrap()` function from the `textwrap` module to solve the problem.
    It includes error handling to check if `n` is a positive integer.

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        List[str]: A list of strings, where each string is a group of `n` consecutive characters from the input string.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Use `textwrap.wrap()` to split the string into groups of `n` characters.
    return textwrap.wrap(s, n)

使用循环和字符串连接:我们还可以通过手动循环字符串并一次将 n 个字符连接到新字符串来解决此问题。一旦我们有 n 个字符,我们就可以将新字符串添加到列表中,并将新字符串重置为空字符串。

def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters.

    This function uses a loop and string concatenation to solve the problem.
    It includes error handling to check if `n` is a positive integer.

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        List[str]: A list of strings, where each string is a group of `n` consecutive characters from the input string.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Initialize an empty list to store the groups.
    result = []

    # Initialize an empty string to store the current group.
    group = ''

    # Iterate over each character in the string.
    for c in s:
        group += c  # Add the current character to the current group.

        # If the current group has `n` characters, add it to the result and reset the group.
        if len(group) == n:
            result.append(group)
            group = ''

    # If there are any remaining characters in the group, add it to the result.
    if group:
        result.append(group)

    return result

使用生成器函数:我们可以创建一个生成器函数,该函数以字符串和数字 n 作为输入,并从字符串中生成 n 个字符的块。这种方法节省内存,因为它不需要一次将所有块存储在内存中。

from typing import Generator


def split_string_into_groups(string: str, n: int) -> Generator[str, None, None]:
    """
    Generator function to split a string into groups of `n` consecutive characters.

    Args:
        string (str): The input string to be split.
        n (int): The size of the groups.

    Yields:
        str: The next group of `n` characters.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> list(split_string_into_groups("HelloWorld", 3))
        ['Hel', 'loW', 'orl', 'd']
        >>> list(split_string_into_groups("Python", 2))
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Iterate over the string with a step size of `n`.
    for i in range(0, len(string), n):
        # Yield the next group of `n` characters.
        yield string[i:i + n]

使用 itertools:Python 中的 itertools 模块提供了一个名为 islice() 的函数,可用于对可迭代对象进行切片。我们可以使用此函数将字符串拆分为 n 个字符的块。

from itertools import islice
from typing import Iterator


def split_string_into_groups(s: str, n: int) -> Iterator[str]:
    """
    Splits a string into groups of `n` consecutive characters using itertools.islice().

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        Iterator[str]: An iterator that yields each group of `n` consecutive characters from the input string.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> list(split_string_into_groups("HelloWorld", 3))
        ['Hel', 'loW', 'orl', 'd']
        >>> list(split_string_into_groups("Python", 2))
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Create an iterator from the string.
    it = iter(s)

    # Use itertools.islice() to yield groups of `n` characters from the iterator.
    while True:
        group = ''.join(islice(it, n))

        if not group:
            break

        yield group

使用 numpy:我们也可以使用 numpy 库来解决这个问题。我们可以将字符串转换为 numpy 数组,然后使用 reshape() 函数将数组拆分为 n 个字符的块。

import numpy as np


def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters using numpy.reshape().

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        List[str]: A list of strings where each string is a group of `n` consecutive characters.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Convert the string to a list of characters
    chars = list(s)

    # Add extra empty strings only if the length of `s` is not a multiple of `n`
    if len(s) % n != 0:
        chars += [''] * (n - len(s) % n)

    # Reshape the array into a 2D array with the number of groups as the number of rows and n as the number of columns
    arr = np.array(chars).reshape(-1, n)

    # Convert each row of the 2D array back to a string and add it to the result list
    result = [''.join(row).rstrip() for row in arr]

    return result

使用 pandas:Python 中的 pandas 库提供了一个名为 groupby() 的函数,可用于将数组拆分为多个 bin。我们可以使用此函数将字符串拆分为 n 个字符的块。

import pandas as pd


def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a given string into groups of `n` consecutive characters.

    This function uses the pandas library to convert the string into a pandas Series,
    then uses the groupby method to group the characters into groups of `n` characters.
    The groups are then converted back to a list of strings.

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        list[str]: A list of strings, where each string is a group of `n` consecutive characters from the input string.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Convert the string to a pandas Series
    s = pd.Series(list(s))

    # Use pandas groupby to group the characters
    # The index of each character is divided by `n` using integer division,
    # which groups the characters into groups of `n` characters.
    groups = s.groupby(s.index // n).agg(''.join)

    # Convert the result back to a list and return it
    return groups.tolist()

使用 more_itertools:more_itertools 库提供了一个名为 chunked() 的函数,可用于将可迭代对象拆分为指定大小的块。我们可以使用此函数将字符串拆分为 n 个字符的块。

import more_itertools


def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters using more_itertools.chunked().

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        List[str]: A list of strings where each string is a group of `n` consecutive characters.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Use more_itertools.chunked() to split the string into chunks of `n` characters.
    chunks = more_itertools.chunked(s, n)

    # Convert each chunk to a string and add it to the result list.
    result = [''.join(chunk) for chunk in chunks]

    return result

使用 toolz:toolz 库提供了一个名为partition_all()的函数,该函数可用于将可迭代对象拆分为指定大小的块。我们可以使用此函数将字符串拆分为 n 个字符的块。

import toolz


def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters using toolz.partition_all().

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        List[str]: A list of strings where each string is a group of `n` consecutive characters.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Use toolz.partition_all() to split the string into chunks of `n` characters.
    chunks = toolz.partition_all(n, s)

    # Convert each chunk to a string and add it to the result list.
    result = [''.join(chunk) for chunk in chunks]

    return result

使用 cytoolz: cytoolz 库提供了一个名为partition_all()的函数,该函数可用于将可迭代对象拆分为指定大小的块。我们可以使用此函数将字符串拆分为n个字符的块。

from cytoolz import partition_all


def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters using cytoolz.partition_all().

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        list[str]: A list of strings where each string is a group of `n` consecutive characters.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Use cytoolz.partition_all() to split the string into chunks of `n` characters.
    chunks = partition_all(n, s)

    # Convert each chunk to a string and add it to the result list.
    result = [''.join(chunk) for chunk in chunks]

    return result

使用 itertools:itertools 库提供了一个名为 zip_longest 的函数,可用于将可迭代对象拆分为指定大小的块。我们可以使用此函数将字符串拆分为 n 个字符的块。

from itertools import zip_longest


def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters using itertools.zip_longest().

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        List[str]: A list of strings where each string is a group of `n` consecutive characters.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Use itertools.zip_longest() to split the string into chunks of `n` characters.
    args = [iter(s)] * n
    chunks = zip_longest(*args, fillvalue='')

    # Convert each chunk to a string and add it to the result list.
    result = [''.join(chunk) for chunk in chunks]

    return result

使用 list + map + join + zip:我们还可以使用 list 函数、map 函数、join 方法和 zip 函数来解决这个问题。我们可以使用 map 函数以步长 n 迭代字符串,并将字符串从当前索引切片到当前索引加 n。然后,我们可以使用 zip 函数将块组合成元组列表,并使用 join 方法将元组连接成字符串列表。

def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters using list, map, join, and zip.

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        list[str]: A list of strings where each string is a group of `n` consecutive characters.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Use list, map, join, and zip to split the string into chunks of `n` characters.
    result = [''.join(chunk) for chunk in zip(*[iter(s)] * n)]

    # If the string length is not a multiple of `n`, add the remaining characters to the result.
    remainder = len(s) % n
    if remainder != 0:
        result.append(s[-remainder:])

    return result

使用递归和切片:我们也可以使用递归和切片来解决这个问题。我们可以定义一个递归函数,该函数以字符串和数字 n 作为输入,并返回一个由 n 个字符组成的块列表。该函数可以将字符串切成 n 个字符的块,并使用剩余的字符串递归调用自身,直到字符串为空。

def split_string_into_groups(s: str, n: int) -> list[str]:
    """
    Splits a string into groups of `n` consecutive characters using recursion with slicing.

    Args:
        s (str): The input string to be split.
        n (int): The size of the groups.

    Returns:
        list[str]: A list of strings where each string is a group of `n` consecutive characters.

    Raises:
        ValueError: If `n` is not a positive integer.

    Examples:
        >>> split_string_into_groups("HelloWorld", 3)
        ['Hel', 'loW', 'orl', 'd']
        >>> split_string_into_groups("Python", 2)
        ['Py', 'th', 'on']
    """
    # Check if `n` is a positive integer.
    if n <= 0:
        raise ValueError("The group size must be a positive integer")

    # Base case: if the length of the string is less than or equal to `n`, return a list containing `s`.
    if len(s) <= n:
        return [s]

    # Recursive case: split the string into two parts and recursively call `split_string_into_groups` on the rest of the string.
    return [s[:n]] + split_string_into_groups(s[n:], n)
相关推荐
  政府信创国产化的10大政策解读一、信创国产化的背景与意义信创国产化,即信息技术应用创新国产化,是当前中国信息技术领域的一个重要发展方向。其核心在于通过自主研发和创新,实现信息技术应用的自主可控,减少对外部技术的依赖,并规避潜在的技术制裁和风险。随着全球信息技术竞争的加剧,以及某些国家对中国在科技领域的打压,信创国产化显...
工程项目管理   1579  
  为什么项目管理通常仍然耗时且低效?您是否还在反复更新电子表格、淹没在便利贴中并参加每周更新会议?这确实是耗费时间和精力。借助软件工具的帮助,您可以一目了然地全面了解您的项目。如今,国内外有足够多优秀的项目管理软件可以帮助您掌控每个项目。什么是项目管理软件?项目管理软件是广泛行业用于项目规划、资源分配和调度的软件。它使项...
项目管理软件   1355  
  信创产品在政府采购中的占比分析随着信息技术的飞速发展以及国家对信息安全重视程度的不断提高,信创产业应运而生并迅速崛起。信创,即信息技术应用创新,旨在实现信息技术领域的自主可控,减少对国外技术的依赖,保障国家信息安全。政府采购作为推动信创产业发展的重要力量,其对信创产品的采购占比情况备受关注。这不仅关系到信创产业的发展前...
信创和国产化的区别   8  
  信创,即信息技术应用创新产业,旨在实现信息技术领域的自主可控,摆脱对国外技术的依赖。近年来,国货国用信创发展势头迅猛,在诸多领域取得了显著成果。这一发展趋势对科技创新产生了深远的推动作用,不仅提升了我国在信息技术领域的自主创新能力,还为经济社会的数字化转型提供了坚实支撑。信创推动核心技术突破信创产业的发展促使企业和科研...
信创工作   9  
  信创技术,即信息技术应用创新产业,旨在实现信息技术领域的自主可控与安全可靠。近年来,信创技术发展迅猛,对中小企业产生了深远的影响,带来了诸多不可忽视的价值。在数字化转型的浪潮中,中小企业面临着激烈的市场竞争和复杂多变的环境,信创技术的出现为它们提供了新的发展机遇和支撑。信创技术对中小企业的影响技术架构变革信创技术促使中...
信创国产化   8  
热门文章
项目管理软件有哪些?
云禅道AD
禅道项目管理软件

云端的项目管理软件

尊享禅道项目软件收费版功能

无需维护,随时随地协同办公

内置subversion和git源码管理

每天备份,随时转为私有部署

免费试用