如何删除字符串中的前导零和尾随零？ Python

Question

我有几个像这样的字母数字字符串

listOfNum = ['000231512-n','1209123100000-n00000','alphanumeric0000', '000alphanumeric']

删除 尾随 零的所需输出将是：

 listOfNum = ['000231512-n','1209123100000-n','alphanumeric', '000alphanumeric']

前导 尾随零的所需输出将是：

 listOfNum = ['231512-n','1209123100000-n00000','alphanumeric0000', 'alphanumeric']

删除前导零和尾随零的期望输出是：

 listOfNum = ['231512-n','1209123100000-n', 'alphanumeric', 'alphanumeric']

现在我一直在做以下方式，如果有请提出更好的方法：

 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)

原文由 alvas 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

基本款怎么样

your_string.strip("0")

删除尾随零和前导零？如果您只对删除尾随零感兴趣，请使用 .rstrip 代替（和 .lstrip 仅用于前导零）。

文档中的更多信息。

您可以使用一些列表理解来获取您想要的序列，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]

原文由 Pierre GM 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

删除前导 + 尾随 ‘0’：

 list = [i.strip('0') for i in listOfNum ]

删除前导“0”：

 list = [ i.lstrip('0') for i in listOfNum ]

删除尾随“0”：

 list = [ i.rstrip('0') for i in listOfNum ]

原文由 fho 发布，翻译遵循 CC BY-SA 3.0 许可协议