如何删除字符串中的前导零和尾随零？python

我有几个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为：

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

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

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

前导尾随零的期望输出为：

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

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

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

删除前导零和尾随零的期望输出为：

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

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

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

目前，我一直采用以下方法，如果有，请建议更好的方法：

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)

那基础教育呢

your_string.strip("0")

要删除尾随零和前导零？如果您只想删除尾随零，请改用

.rstrip

（而

.lstrip

仅用于前导零）

更多信息请访问

您可以使用一些列表理解来获得您想要的序列，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]

您是否尝试过：

---

删除前导+尾随“0”：

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

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

删除前导“0”：

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

删除尾部“0”：

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

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

---

您只需使用bool即可完成此操作：

if int(number) == float(number):

   number = int(number)

else:

   number = float(number)

---

str.strip

是解决这种情况的最佳方法，但也是一种通用解决方案，它可以从iterable中剥离前导元素和尾随元素：

代码

import more_itertools as mit


iterables = ["231512-n\n","  12091231000-n00000","alphanum0000", "00alphanum"]
pred = lambda x: x in {"0", "\n", " "}
list("".join(mit.strip(i, pred)) for i in iterables)
# ['231512-n', '12091231000-n', 'alphanum', 'alphanum']

---

详细信息

注意，在这里，我们去掉了满足谓词的其他元素中的前导和尾随

“0”

s。此工具不限于字符串

另请参见文档以了解更多的示例

• ：两端剥开
• ：剥离左端
• ：剥去右端

---

第三方库是否可通过

>pip安装更多\u itertools

安装？如果列表中有其他数据类型（而不仅仅是字符串），请尝试此操作。这将从字符串中删除尾随零和前导零，并保留其他数据类型不变。这也会处理特殊情况s='0'

e、 g

---

不能按照OP的要求使用

alphanumeric0000

。对于

s='0'

的特殊情况，这个答案有没有巧妙的调整？@Charles:是的！我刚才遇到了同样的问题，您可以执行

s.strip（“0”）或“0”

：如果您的字符串变为空字符串，它将通过或计算为

False

，并将替换为所需的字符串

“0”

@Pierre，谢谢。这真的很有帮助，而且非常简单。高估