I want to get only the numbers from a string. For example I have something like this
just='Standard Price:20000'
And I only want it to print out
20000
So I can multiply it by any number.
I tried
just='Standard Price:20000'
just[0][1:]
I got
''
What's the best way to go about this? I'm a noob.
you can use regex:
import re
just = 'Standard Price:20000'
price = re.findall("\d+", just)[0]
OR
price = just.split(":")[1]
You can also try:
int(''.join(i for i in just if i.isdigit()))
If you want to keep it simpler avoiding regex, you can also try Python's built-in function filter with str.isdigit function to get the string of digits and convert the returned string to integer. This will not work for float as the decimal character is filtered out by str.isdigit.
Python Built-in Functions Filter
Python Built-in Types str.isdigit
Considering the same code from the question:
>>> just='Standard Price:20000'
>>> price = int(filter(str.isdigit, just))
>>> price
20000
>>> type(price)
<type 'int'>
>>>
I think bdev TJ's answer
price = int(filter(str.isdigit, just))
will only work in Python2, for Python3 (3.7 is what I checked) use:
price = int ( ''.join(filter(str.isdigit, just) ) )
Obviously and as stated before, this approach will only yield an integer containing all the digits 0-9 in sequence from an input string, nothing more.
You could use string.split function.
>>> just='Standard Price:20000'
>>> int(just.split(':')[1])
20000
The clearest way in my opinion is using the re.sub() function:
import re
just = 'Standard Price:20000'
only_number = re.sub('[^0-9]', '', just)
print(only_number)
# Result: '20000'
You could use RegEx
>>> import re
>>> just='Standard Price:20000'
>>> re.search(r'\d+',just).group()
'20000'
Ref: \d matches digits from 0 to 9
just[0] evaluates to S as it is the 0th character. Thus S[1:] returns an empty string that is '', because the string is of length 1 and there are no other characters after length 1
A more crude way if you don't want to use regex is to make use of slicing. Please remember that this would work for extracting any number if the structure of the text remains the same.
just = 'Standard Price:20000'
reqChar = len(just) - len('Standard Price:')
print(int(just[-reqChar:]))
>> 20000
Not much difference in what you use but avoiding regex saves an extra import (i5-8250U).
import re
def regex():
dirty = 'Standard Price:20000'
return re.findall("\\d+", dirty)[0]
def regex_grp():
dirty = 'Standard Price:20000'
return re.search(r'\d+', dirty).group()
def isdigit():
dirty = 'Standard Price:20000'
return int(''.join(i for i in dirty if i.isdigit()))
def filter_():
dirty = 'Standard Price:20000'
return int(''.join(filter(str.isdigit, dirty)))
regex() 1.33 μs ± 17.2 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each)
regex_grp() 1.4 μs ± 17.9 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each)
isdigit() 1.73 μs ± 5.73 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each)
filter_() 1.33 μs ± 23.1 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each)
just[0] == 'S',right?"".join(re.findall('\d+', just))to get all numbers from string