Python:如何拆分()每个符号,直到到达分隔符符号



我有一个字符串:

a = "12356789:10:51:52:53"

因此,我需要一个列表,如:

['1','2','3','5','6','7','8','9','10','51','52','53']

如何逐个拆分符号,直到找到:分隔符?

你可以试试这个:

list((x:="12356789:10:51:52:53".split(":"))[0]) + x[1:]

一种简短而棘手的娱乐方式(不建议用于生产代码:-(:

a = "12356789:10:51:52:53"
[*b], *b[len(a):] = a.split(':')
print(b)

输出:

['1', '2', '3', '5', '6', '7', '8', '9', '10', '51', '52', '53']

这样行吗?

a = "12356789:10:51:52:53"
l1 = a.split(":")
l2 = [c for c in l1[0]]
output = l2 + l1[1:]
output 
In [193]: a = "12356789:10:51:52:53"
In [194]: first_split = a.index(':')
In [195]: a[:first_split].split()
Out[195]: ['12356789']
In [196]: a = "12356789:10:51:52:53"
In [197]: b = list(a[:first_split])
Out[197]: ['1', '2', '3', '5', '6', '7', '8', '9']
In [198]: c = a[first_split+1:].split(':')
Out[198]: ['10', '51', '52', '53']

然后可以组合b+c

不确定这是否是最好的解决方案,但这会起作用:

a = "12356789:10:51:52:53"
list(''.join(a.split(":")[0])) + a.split(":")[1:]

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