See if any characters are in string and remove them

Question

So I have a simple program in Python 3 that is just a basic calculator. here is a section of code:

calculation = input("What calculation do you want to do?\n")
if "+" in calculation:
    numbers = calculation.split("+")
    answer = int(numbers[0]) + int(numbers[1])

I have some other operations setup below it. The issue is that if someone was to input anything as well as the operation (e.g. 10++2 or 10+abc2), the code throws back an error because it obviously can't add 10 to abc2.

I thought that I could solve the issue by testing for each individual character but surely that is a long way around the issue. Is there a way to solve the issue using Python?

This is somewhat similar to a [question](http://stackoverflow.com/q/31999444/892383) I recently [answered](http://stackoverflow.com/a/32022459/892383). — Cyphase, Aug 16 '15 at 11:25

Assem · Accepted Answer · 2015-08-16T12:00:05.937

5

Check if both sides are made of digits by the string.isdigit() method:

if numbers[0].isdigit() and numbers[1].isdigit():
     answer = int(numbers[0]) + int(numbers[1])

If you want to go further with your calculator, you should worry about learning regex and parsing.

edited Aug 16 '15 at 12:00

answered Aug 16 '15 at 11:26

Assem

11,574
5
59
97

This will fail for `2.0` etc.. which is a valid number – Padraic Cunningham Aug 16 '15 at 11:53
I understood that OP is just starting a basic calculator that perform `int+int`. I updated the answer to refer that he should go with parsing and regex for more complicated operations. – Assem Aug 16 '15 at 12:02

score 3 · Answer 2 · answered Aug 16 '15 at 11:25

3

Python, you love it too?

s = ''.join(filter(lambda x: x.isdigit() or x=='+', list(s)))

answered Aug 16 '15 at 11:25

xrisk

3,790
22
45

1

The author wanted to throw error in case there is invalid character, but ```filter``` will just ignore invalid parts. I'd use ```if not all(lambda x: x.isdigit() or x=='+', s)``` if I wanted to make sure the input is valid. Also note that there is no need to convert ```s``` to ```list```, as ```str``` is iterable – tmrlvi Aug 16 '15 at 11:35
@tmrlvi feel free to suggest an edit! I’m just a beginner in python. – xrisk Aug 16 '15 at 11:37

score 1 · Answer 3 · 2015-08-16T12:19:05.580

1

Try this:

calculation = "5+4+3.5"

numbers = calculation.split("+")
numbers2 = []

for num in numbers:
    try:
        numbers2.append(float(num))
    except:
        print("incorrect input")

print(sum(numbers2))

edited Aug 16 '15 at 12:19

answered Aug 16 '15 at 11:28

1

```num``` is a string, so ```isinstance(num, (int, long, float, complex))``` will always return ```False``` – tmrlvi Aug 16 '15 at 11:29
1

Now ```4.5``` would result in ```"incorrect input"``` (so the last statement is incorrect) – tmrlvi Aug 16 '15 at 11:36
`false` is not python, you can also `if not num.isdigit()` , your code will also fail for multiple different inputs – Padraic Cunningham Aug 16 '15 at 11:56
Ok, I fixed it guys... I think the general approach was understood though. – Aug 16 '15 at 12:20

Padraic Cunningham · Answer 4 · 2015-08-16T11:58:54.960

You could use a regex to get the digits:

s = "10++abc2.0"

import re
if "+" in s:
    print(sum(map(float,re.findall("\d+|\d+\.\d+", s))))

12.0

If you want to ignore abc2.0 and catch negative numbers, try casting with a try/except:

def parse(s, sign):
    spl = s.split(sign)
    sm = 0
    for i in spl:
        try:
            print(i)
            sm += float(i)
        except ValueError:
            pass
    return sm

sm = parse(s, "+")
print(sm)
10.0

You have to also consider negative numbers but there are cases where you could have + and - in the string so you also need to decide what do do then:

s = "10++abc2.0+-2"
sm = parse(s,"+")
print(sm)
8.0

See if *any* characters are in string and remove them

4 Answers4

See if any characters are in string and remove them