How can I remove digits from a string?
78 Answers
Would this work for your situation?
>>> 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:
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 :) :
>>> s = '12abcd405' >>> result = ''.join(i for i in s if not i.isdigit()) >>> result 'abcd' And, just to throw it in the mix, is the oft-forgotten str.translate which will work a lot faster than looping/regular expressions:
For Python 2:
from string import digits s = 'abc123def456ghi789zero0' res = s.translate(None, digits) # 'abcdefghizero' For Python 3:
from string import digits s = 'abc123def456ghi789zero0' remove_digits = str.maketrans('', '', digits) res = s.translate(remove_digits) # 'abcdefghizero' Not sure if your teacher allows you to use filters but...
filter(lambda x: x.isalpha(), "a1a2a3s3d4f5fg6h") returns-
'aaasdffgh' Much more efficient than looping...
Example:
for i in range(10): a.replace(str(i),'') What about this:
out_string = filter(lambda c: not c.isdigit(), in_string) Just a few (others have suggested some of these)
Method 1:
''.join(i for i in myStr if not i.isdigit()) Method 2:
def removeDigits(s): answer = [] for char in s: if not char.isdigit(): answer.append(char) return ''.join(answer) Method 3:
''.join(filter(lambda x: not x.isdigit(), mystr)) Method 4:
nums = set(map(int, range(10))) ''.join(i for i in mystr if i not in nums) Method 5:
''.join(i for i in mystr if ord(i) not in range(48, 58)) Say st is your unformatted string, then run
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 s is your string and st_res is a string without digits, then here is your code
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 I'd love to use regex to accomplish this, but since you can only use lists, loops, functions, etc..
here's what I came up with:
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! 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
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