Removing numbers from string [closed]

Question

How can I remove digits from a string?

Percolator

With re: result = re.sub(r'[0-9]+', '', s) – Notochord

with regex you will need to add \. also, as it can be decimal number i think. like result = re.sub(r'[0-9\.]+', '', s) – Phosphorus

"\d" is the same in a regex as "[0-9]", so you can do result = re.sub(r"\d+", "", s) instead. Speed will probably depend on the particular string being used, but for me, re.sub took about twice as long as str.translate (slightly longer if you don't use a pre-compiled pattern). – Facelift

@WiktorStribiżew, you answer is working fine but it is adding a new line in the file. Any reason? – Salzhauer

@LakshmiYadav re.sub(r'[0-9]+', '', s) removes found matches (see the second argument that is an empty string), it can't add anything. Check your code. – Notochord

@WiktorStribiżew, I have 4 line of numeric and alphanumeric strings in a file, I am using for loop and adding your line of code. import fileinput import re for line in fileinput.input("/Users/xyz/Desktop/temp/i_tmp.txt", inplace=True): print re.sub(r'\b[0-9\.]+','', line) Once I run above code, numbers are vanished but after every line new line is been added. – Salzhauer

@LakshmiYadav It has nothing to do with my regex. Check your print. – Notochord

This should instead be a duplicate of the much better-asked stackoverflow.com/questions/15754587. – Cutcheon

Answer 1

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'

Blackandblue

@SeanJohnson Awesome! I'm sure I learned that from somebody else on this site, so the cycle is complete :) – Blackandblue

@Blackandblue no need of [] – Psilomelane

In Python 2.7 and above, you don't need the brackets around the list comprehension. You can leave them out and it becomes a generator expression. – Fisken

@Blackandblue add some explanation too, just seeing the code won't help the OP I guess. – Psilomelane

@AshwiniChaudhary Good point - added (put the list comprehension version first to make the for-loop comparison a bit easier to decipher). – Blackandblue

Jon Clements answer below using s.translate is about 10 times faster than the list comprehension (tested with Python 2.7). Just a heads up to anyone reading this later. – Natch

Answer 2

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'

Pronuba

This approach won't work in Python3. Do instead: 'abc123def456ghi789zero0'.translate({ord(k): None for k in digits}) – Jelly

Best solution for Python2. – Gapeworm

Doesn't work for unicode character strings – Komsa

Answer 3

Not sure if your teacher allows you to use filters but...

"".join(filter(lambda x: x.isalpha(), "a1a2a3s3d4f5fg6h"))

returns-

'aaasdffgh'

Much more efficient than looping...

Example:

for i in range(10):
  a.replace(str(i),'')

Chelsea

it returns this instead : <filter object at 0x03475FD0> – Scenery

You can get a sting back by using "".join(filter(lambda x: x.isalpha(), "a1a2a3s3d4f5fg6h")). That's an empty string in the beginning. – Adda

Or better: "".join(filter(lambda ch: not ch.isdigit(), "a1a2a3s3d4f5fg6h")) --- so not isdigit is better suits the question. (I can't edit the answer :-( There are too many "edit is awaiting approval" message in the last half year.) – Parabasis

Answer 4

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))

Zymogen

It would be worthwhile to show efficiency comparison on these. – Ute

In method 2, it should be ''.join(answer) not ''.join(char). – Hae

@Sreram: you're absolutely right! Thanks for the bugreport – Zymogen

Answer 5

What about this:

out_string = filter(lambda c: not c.isdigit(), in_string)

Knowledgeable

Output is <filter object at 0x7f749e1745c0>. Python3.6 – Allowed

@Allowed You can coerce the generator into a list object by wrapping that returned object from filter into list(filter(...)) – Steelworks

Answer 6

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

Alcine

Answer 7

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!

Megalomania

Answer 8

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

Sheelah

A for-loop automatically iterates through every character of a string, so no need of breaking the string into chars. – Psilomelane