英文:
If the string contains an odd number of characters, how to add a ( _ ) character after the last character
问题
在最后一个字符后添加一个下划线(_)的方法是什么？
def split_string(string: str) -> list:n = 2list1 = [string[i: i + n] for i in range(0, len(string), n)]return list1print(split_string("123456")) # ["12", "34", "56"]print(split_string("ab cd ef")) # ["ab", " c", "d ", "ef"]print(split_string("abc")) # ["ab", "c_"]print(split_string(" ")) # [" _"]print(split_string("")) # []
英文:
How to add a character(_) after the last character?
def split_string(string: str) -> list:n = 2list1 = [string[i: i + n] for i in range(0, len(string), n)]return list1print(split_string("123456")) # ["12", "34", "56"]print(split_string("ab cd ef")) # ["ab", " c", "d ", "ef"]print(split_string("abc")) # ["ab", "c_"]print(split_string(" ")) # [" _"]print(split_string("")) # []
答案1
得分: 1
只需向临时副本添加一个字符，然后使用该字符进行操作：
def split_string(string: str) -> list:s = string + "_" # <----list1 =展开收缩for i in range(0, len(string), 2)]# ^return list1
英文:
Just add a character to a temporary copy, and work with that:
def split_string(string: str) -> list:s = string + "_" # <----list1 =展开收缩for i in range(0, len(string), 2)]# ^return list1
答案2
得分: 1
itertools 模块提供了 grouper 函数的定义：
from itertools import zip_longestdef grouper(iterable, n, *, incomplete='fill', fillvalue=None):"将数据收集到非重叠的固定长度块中"# grouper('ABCDEFG', 3, fillvalue='x') --> ABC DEF Gxx# grouper('ABCDEFG', 3, incomplete='strict') --> ABC DEF ValueError# grouper('ABCDEFG', 3, incomplete='ignore') --> ABC DEFargs = [iter(iterable)] * nif 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')
你的函数是一个特殊情况：它将字符串分割成长度为2的组，使用“_”填充不完整的块，并将结果的2元组连接成单个字符串。
def split_string(s):return list(map(''.join, grouper(s, 2, fillvalue="_")))# return [''.join(t) for t in grouper(s, 2, fillvalue="_")]
英文:
The itertools module provides the definition of a function grouper:
from itertools import zip_longestdef 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 DEFargs = [iter(iterable)] * nif 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')
Your function is a special case: you split your string into groups of 2, padding incomplete chunks with "_", and joining the resulting 2-tuples back into a single string.
def split_string(s):return list(map(''.join, grouper(s, 2, fillvalue="_")))# return [''.join(t) for t in grouper(s, 2, fillvalue="_")]
答案3
得分: 1
你可以在返回列表之前调整最后一个元素：
if len(string) % 2:list1[-1] += "_"
或者支持其他值的 n：
if add := -len(string) % n:list1[-1] += "_" * add
或者，使用迭代器：
def split_string(string: str) -> list:it = iter(string)return [c + next(it, "_") for c in it]
英文:
You could just adjust the last element before you return the list:
if len(string) % 2:list1[-1] += "_"
Or to support other values of n:
if add := -len(string) % n:list1[-1] += "_" * add
Alternatively, with an iterator:
def split_string(string: str) -> list:it = iter(string)return [c + next(it, "_") for c in it]
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