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I am trying to figure out a way to detect a string input of let's say 54% and convert that to decimal. I can get the percentage to convert with a simple if() statement and multiplying it by a 100. I just can't figure out what I should set the if statements argument to?

dfundako
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Mason Swanner
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  • Can you share your code even if it does not work correctly? Also, you can add sample inputs and outputs for your program. – Alperen Jul 20 '20 at 19:10
  • `"%" == "%"`, although if you want to print 50% as `"%d %" % 50` it will raise ValueError you have to use `"%d %%" % 50` – Grijesh Chauhan Jul 20 '20 at 19:11
  • Does this answer your question? [Replacing instances of a character in a string](https://stackoverflow.com/questions/12723751/replacing-instances-of-a-character-in-a-string) – Roim Jul 20 '20 at 19:15
  • Does this answer your question? [How to check a string for specific characters?](https://stackoverflow.com/questions/5188792/how-to-check-a-string-for-specific-characters) – Gino Mempin Jul 21 '20 at 00:07

3 Answers3

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lst = ['54%', '54', '0.37%', 'apple%']

# Go through the list
for i in lst:

    # Check whether there's a percent sign
    if '%' in i:

        # Try whether only taking away the % sign is enough for it to be able to be divideable
        try:

            # Take out that % sign
            percent = float(i.replace('%', ''))
            decimal = percent / 100
            print(decimal)

        # If taking the % sign away is not enough, do the following stuff
        except:

            # Create an empty list which we will append to later and set a variable to 0 
            lstnumbers = []
            errors = 0

            # Go through every character of the item
            for numbers in i:

                # Check whether it can be changed to an integer and append it to a list
                try:
                    number = int(numbers)
                    lstnumbers.append(number)

                # Increase the variable by 1 if there's a charachter that can't be converted to an integer
                except:
                    errors += 1

            # If there's something with a % that can't be converted to an integer, print that
            if errors > 0:
                print("there's a value with a % sign, which cannot be converted:", i)

            # If there were some integers in the item, do something
            if len(lstnumbers) > 0:

                # Replace the first 0 as it the input could also be 0.46 for example and 046 would be no number we'd be interested in
                lstnumbersnew = ['' if lstnumbers[0] == '0' else number for number in lstnumbers]

                # Bring the items from the list together so we have string
                finalnumber = ''.join(str(lstnumbersnew))

                # Make that string an integer and divide it
                finalintegers = int(float(finalnumber)) / 100

                # Print it or do whatever you want to do with it
                print(finalintegers)

This will output:

0.54
0.0037
there's a value with a % sign, which cannot be converted: apple%
Jem
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  • ```percent = int(i.replace('%', ''))``` this will break if there are other characters present, like in '.54%' – M-Wi Jul 20 '20 at 19:20
  • @M-Wi You were absolutely right, my previous code was very easy to break. It took some time but I enhanced my code which should make it capable handling a lot more diverse input. – Jem Jul 20 '20 at 19:49
0
input = input(': ')
percent = (% in input)
if percent == True:
    #Do what you want
    print('Correct')
else:
    print('Wrong')
Plum The Coder
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0

You can use an in line if like this:

def detectInput(inp, charToDetect): # inp = the user input
    # if % in inp then convert else do nothing
    print(((int(inp.replace(charToDetect, '')))/100) if (charToDetect in inp) else "")

x = "54%"
detectInput(x, '%')
tibi
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