# Removing numbers from string

## Question:

How can I remove digits from a string?

## Answers:

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!
```

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 ðŸ™‚ :

```
>>> s = '12abcd405'
>>> result = ''.join(i for i in s if not i.isdigit())
>>> result
'abcd'
```

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

What about this:

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

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
```

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),'')
```

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))
```

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'
```