我有几个像这样的字母数字字符串
listOfNum = ['000231512-n','1209123100000-n00000','alphanumeric0000', '000alphanumeric']
除去 尾随 零的理想输出为：
listOfNum = ['000231512-n','1209123100000-n','alphanumeric', '000alphanumeric']
前导 尾随零的期望输出为：
listOfNum = ['231512-n','1209123100000-n00000','alphanumeric0000', 'alphanumeric']
去除前导零和尾随零的期望输出为：
listOfNum = ['231512-n','1209123100000-n', 'alphanumeric', 'alphanumeric']
目前，我已经按照以下方式进行操作，如果有的话，请提出一种更好的方法：
listOfNum = ['000231512-n','1209123100000-n00000','alphanumeric0000', \ '000alphanumeric'] trailingremoved = [] leadingremoved = [] bothremoved = [] # Remove trailing for i in listOfNum: while i[-1] == "0": i = i[:-1] trailingremoved.append(i) # Remove leading for i in listOfNum: while i[0] == "0": i = i[1:] leadingremoved.append(i) # Remove both for i in listOfNum: while i[0] == "0": i = i[1:] while i[-1] == "0": i = i[:-1] bothremoved.append(i)
那基本的
your_string.strip("0")
删除尾随和前导零？如果您只想删除尾随零，请.rstrip改用（.lstrip仅用于前导零）。
.rstrip
.lstrip
[文档中的更多信息。]
您可以使用一些列表推导来获得所需的序列，如下所示：
trailing_removed = [s.rstrip("0") for s in listOfNum] leading_removed = [s.lstrip("0") for s in listOfNum] both_removed = [s.strip("0") for s in listOfNum]
