问题：如何删除字符串中的前导和尾随零？Python
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']
除去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)
I have several alphanumeric strings like these
listOfNum = ['000231512-n','1209123100000-n00000','alphanumeric0000', '000alphanumeric']
The desired output for removing trailing zeros would be:
listOfNum = ['000231512-n','1209123100000-n','alphanumeric', '000alphanumeric']
The desired output for leading trailing zeros would be:
listOfNum = ['231512-n','1209123100000-n00000','alphanumeric0000', 'alphanumeric']
The desire output for removing both leading and trailing zeros would be:
listOfNum = ['231512-n','1209123100000-n', 'alphanumeric', 'alphanumeric']
For now i’ve been doing it the following way, please suggest a better way if there is:
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)
回答 0
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）。
[ 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]
What about a basic
your_string.strip("0")
to remove both trailing and leading zeros ? If you’re only interested in removing trailing zeros, use .rstrip instead (and .lstrip for only the leading ones).
[More info in the doc.]
You could use some list comprehension to get the sequences you want like so:
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]
回答 1
删除前导+尾随的“ 0”：
list = [i.strip('0') for i in listOfNum ]
删除前导“ 0”：
list = [ i.lstrip('0') for i in listOfNum ]
删除尾随的“ 0”：
list = [ i.rstrip('0') for i in listOfNum ]
Remove leading + trailing ‘0’:
list = [i.strip('0') for i in listOfNum ]
Remove leading ‘0’:
list = [ i.lstrip('0') for i in listOfNum ]
Remove trailing ‘0’:
list = [ i.rstrip('0') for i in listOfNum ]
回答 2
您可以简单地通过bool做到这一点：
if int(number) == float(number):
number = int(number)
else:
number = float(number)
You can simply do this with a bool:
if int(number) == float(number):
number = int(number)
else:
number = float(number)
回答 3
您是否尝试了strip（）：
listOfNum = ['231512-n','1209123100000-n00000','alphanumeric0000', 'alphanumeric']
print [item.strip('0') for item in listOfNum]
>>> ['231512-n', '1209123100000-n', 'alphanumeric', 'alphanumeric']
Did you try with strip() :
listOfNum = ['231512-n','1209123100000-n00000','alphanumeric0000', 'alphanumeric']
print [item.strip('0') for item in listOfNum]
>>> ['231512-n', '1209123100000-n', 'alphanumeric', 'alphanumeric']
回答 4
str.strip是解决这种情况的最佳方法，但more_itertools.strip还是一种通用解决方案，可从迭代中剥离前导元素和尾随元素：
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']
细节
注意，这里我们"0"将满足谓词的其他元素中的前导和尾随s 剥离。此工具不仅限于字符串。
另请参阅docs，以获取更多的示例
more_itertools是可通过安装的第三方库> pip install more_itertools。
str.strip is the best approach for this situation, but more_itertools.strip is also a general solution that strips both leading and trailing elements from an iterable:
Code
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']
Details
Notice, here we strip both leading and trailing "0"s among other elements that satisfy a predicate. This tool is not limited to strings.
See also docs for more examples of
more_itertools is a third-party library installable via > pip install more_itertools.
回答 5
假设列表中还有其他数据类型（不仅是字符串），请尝试此操作。这将从字符串中删除尾随和前导零，并使其他数据类型保持不变。这也处理特殊情况s =’0′
例如
a = ['001', '200', 'akdl00', 200, 100, '0']
b = [(lambda x: x.strip('0') if isinstance(x,str) and len(x) != 1 else x)(x) for x in a]
b
>>>['1', '2', 'akdl', 200, 100, '0']
Assuming you have other data types (and not only string) in your list try this. This removes trailing and leading zeros from strings and leaves other data types untouched. This also handles the special case s = ‘0’
e.g
a = ['001', '200', 'akdl00', 200, 100, '0']
b = [(lambda x: x.strip('0') if isinstance(x,str) and len(x) != 1 else x)(x) for x in a]
b
>>>['1', '2', 'akdl', 200, 100, '0']
