Python 从字符串中删除数字
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Removing numbers from string
提问by user1739954
How can I remove digits from a string?
如何从字符串中删除数字？
采纳答案by RocketDonkey
Would this work for your situation?
这对你的情况有用吗？
>>> s = '12abcd405'
>>> result = ''.join([i for i in s if not i.isdigit()])
>>> result
'abcd'
This makes use of a list comprehension, and what is happening here is similar to this structure:
这使用了列表推导式，这里发生的事情类似于这个结构：
no_digits = []
# Iterate through the string, adding non-numbers to the no_digits list
for i in s:
if not i.isdigit():
no_digits.append(i)
# Now join all elements of the list with '',
# which puts all of the characters together.
result = ''.join(no_digits)
As @AshwiniChaudhary and @KirkStrauser point out, you actually do not need to use the brackets in the one-liner, making the piece inside the parentheses a generator expression (more efficient than a list comprehension). Even if this doesn't fit the requirements for your assignment, it is something you should read about eventually :) :
正如@AshwiniChaudhary 和@KirkStrauser 指出的那样，您实际上不需要在单行中使用括号，从而使括号内的部分成为生成器表达式（比列表理解更有效）。即使这不符合您的任务要求，您最终也应该阅读它:)：
>>> s = '12abcd405'
>>> result = ''.join(i for i in s if not i.isdigit())
>>> result
'abcd'
回答by Sean Johnson
I'd love to use regex to accomplish this, but since you can only use lists, loops, functions, etc..
我很想使用正则表达式来完成此操作，但由于您只能使用列表、循环、函数等。
here's what I came up with:
这是我想出的：
stringWithNumbers="I have 10 bananas for my 5 monkeys!"
stringWithoutNumbers=''.join(c if c not in map(str,range(0,10)) else "" for c in stringWithNumbers)
print(stringWithoutNumbers) #I have bananas for my monkeys!
回答by Baahubali
If i understand your question right, one way to do is break down the string in chars and then check each char in that string using a loop whether it's a string or a number and then if string save it in a variable and then once the loop is finished, display that to the user
如果我理解你的问题是正确的，一种方法是将字符串分解为字符，然后使用循环检查该字符串中的每个字符是字符串还是数字，然后如果字符串将其保存在变量中，然后进行一次循环完成，显示给用户
回答by Pavel Paulau
What about this:
那这个呢：
out_string = filter(lambda c: not c.isdigit(), in_string)
回答by iddqd
Say st is your unformatted string, then run
说 st 是你的未格式化字符串，然后运行
st_nodigits=''.join(i for i in st if i.isalpha())
as mentioned above. But my guess that you need something very simple so say sis your string and st_resis a string without digits, then here is your code
正如刚才提到的。但我猜你需要一些非常简单的东西，所以说s是你的字符串，st_res是一个没有数字的字符串，那么这里是你的代码
l = ['0','1','2','3','4','5','6','7','8','9']
st_res=""
for ch in s:
if ch not in l:
st_res+=ch
回答by Alain Nisam
Not sure if your teacher allows you to use filters but...
不确定您的老师是否允许您使用过滤器，但是...
filter(lambda x: x.isalpha(), "a1a2a3s3d4f5fg6h")
returns-
返回-
'aaasdffgh'
Much more efficient than looping...
比循环更有效...
Example:
例子：
for i in range(10):
a.replace(str(i),'')
回答by inspectorG4dget
Just a few (others have suggested some of these)
只是一些（其他人已经建议了其中的一些）
Method 1:
方法一：
''.join(i for i in myStr if not i.isdigit())
Method 2:
方法二：
def removeDigits(s):
answer = []
for char in s:
if not char.isdigit():
answer.append(char)
return ''.join(char)
Method 3:
方法三：
''.join(filter(lambda x: not x.isdigit(), mystr))
Method 4:
方法四：
nums = set(map(int, range(10)))
''.join(i for i in mystr if i not in nums)
Method 5:
方法五：
''.join(i for i in mystr if ord(i) not in range(48, 58))
回答by Jon Clements
And, just to throw it in the mix, is the oft-forgotten str.translatewhich will work a lot faster than looping/regular expressions:
而且，只是把它放在混合中，是经常被遗忘的str.translate，它比循环/正则表达式的工作速度要快得多：
For Python 2:
对于 Python 2：
from string import digits
s = 'abc123def456ghi789zero0'
res = s.translate(None, digits)
# 'abcdefghizero'
For Python 3:
对于 Python 3：
from string import digits
s = 'abc123def456ghi789zero0'
remove_digits = str.maketrans('', '', digits)
res = s.translate(remove_digits)
# 'abcdefghizero'
