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:]