我想从字符串中删除除第一个字符以外的所有特定字符匹配。
示例代码
specific_char = ','
example_str = '110,49144,35,123'
# Apply here magical one-liner.
print(example_str) # '110,4914435123'
谢谢!
答案 0 :(得分:4)
也许这
specific_char = ','
x = example_str.split(specific_char)
x[0]+specific_char+''.join(x[1:])
答案 1 :(得分:-1)
只是为了好玩:
example_str = example_str[::-1].replace(",", "", example_str.count(",")-1)[::-1]
答案 2 :(得分:-1)
您可以使用enumerate
:
c = ','
s = '110,49144,35,123'
result = ''.join(a for i, a in enumerate(s) if a != c or c not in s[:i])
输出:
'110,4914435123'
答案 3 :(得分:-2)
这就是你可以做的
specific_char = ','
example_str = '110,49144,35,123'
index = 0
for i in range(len(example_str)):
if example_str[i] == specific_char:
index = i
break
example_str = example_str[:index+1]+example_str[index:].replace(specific_char,'')
输出
'110,4914435123'