How to extract numbers from a string in Python?
Question:
I would like to extract all the numbers contained in a string. Which is better suited for the purpose, regular expressions or the isdigit() method?
Example:
line = "hello 12 hi 89"
Result:
[12, 89]
Answers:
I’d use a regexp :
>>> import re
>>> re.findall(r'd+', "hello 42 I'm a 32 string 30")
['42', '32', '30']
This would also match 42 from bla42bla. If you only want numbers delimited by word boundaries (space, period, comma), you can use b :
>>> re.findall(r'bd+b', "he33llo 42 I'm a 32 string 30")
['42', '32', '30']
To end up with a list of numbers instead of a list of strings:
>>> [int(s) for s in re.findall(r'bd+b', "he33llo 42 I'm a 32 string 30")]
[42, 32, 30]
NOTE: this does not work for negative integers
I’m assuming you want floats not just integers so I’d do something like this:
l = []
for t in s.split():
try:
l.append(float(t))
except ValueError:
pass
Note that some of the other solutions posted here don’t work with negative numbers:
>>> re.findall(r'bd+b', 'he33llo 42 I'm a 32 string -30')
['42', '32', '30']
>>> '-3'.isdigit()
False
If you only want to extract only positive integers, try the following:
>>> txt = "h3110 23 cat 444.4 rabbit 11 2 dog"
>>> [int(s) for s in txt.split() if s.isdigit()]
[23, 11, 2]
I would argue that this is better than the regex example because you don’t need another module and it’s more readable because you don’t need to parse (and learn) the regex mini-language.
This will not recognize floats, negative integers, or integers in hexadecimal format. If you can’t accept these limitations, jmnas’s answer below will do the trick.
@jmnas, I liked your answer, but it didn’t find floats. I’m working on a script to parse code going to a CNC mill and needed to find both X and Y dimensions that can be integers or floats, so I adapted your code to the following. This finds int, float with positive and negative vals. Still doesn’t find hex formatted values but you could add “x” and “A” through “F” to the num_char tuple and I think it would parse things like ‘0x23AC’.
s = 'hello X42 I'm a Y-32.35 string Z30'
xy = ("X", "Y")
num_char = (".", "+", "-")
l = []
tokens = s.split()
for token in tokens:
if token.startswith(xy):
num = ""
for char in token:
# print(char)
if char.isdigit() or (char in num_char):
num = num + char
try:
l.append(float(num))
except ValueError:
pass
print(l)
This is more than a bit late, but you can extend the regex expression to account for scientific notation too.
import re
# Format is [(<string>, <expected output>), ...]
ss = [("apple-12.34 ba33na fanc-14.23e-2yapple+45e5+67.56E+3",
['-12.34', '33', '-14.23e-2', '+45e5', '+67.56E+3']),
('hello X42 I'm a Y-32.35 string Z30',
['42', '-32.35', '30']),
('he33llo 42 I'm a 32 string -30',
['33', '42', '32', '-30']),
('h3110 23 cat 444.4 rabbit 11 2 dog',
['3110', '23', '444.4', '11', '2']),
('hello 12 hi 89',
['12', '89']),
('4',
['4']),
('I like 74,600 commas not,500',
['74,600', '500']),
('I like bad math 1+2=.001',
['1', '+2', '.001'])]
for s, r in ss:
rr = re.findall("[-+]?[.]?[d]+(?:,ddd)*[.]?d*(?:[eE][-+]?d+)?", s)
if rr == r:
print('GOOD')
else:
print('WRONG', rr, 'should be', r)
Gives all good!
Additionally, you can look at the AWS Glue built-in regex
The best option I found is below. It will extract a number and can eliminate any type of char.
def extract_nbr(input_str):
if input_str is None or input_str == '':
return 0
out_number = ''
for ele in input_str:
if ele.isdigit():
out_number += ele
return float(out_number)
This answer also contains the case when the number is float in the string
def get_first_nbr_from_str(input_str):
'''
:param input_str: strings that contains digit and words
:return: the number extracted from the input_str
demo:
'ab324.23.123xyz': 324.23
'.5abc44': 0.5
'''
if not input_str and not isinstance(input_str, str):
return 0
out_number = ''
for ele in input_str:
if (ele == '.' and '.' not in out_number) or ele.isdigit():
out_number += ele
elif out_number:
break
return float(out_number)
If you know it will be only one number in the string, i.e 'hello 12 hi', you can try filter.
For example:
In [1]: int(''.join(filter(str.isdigit, '200 grams')))
Out[1]: 200
In [2]: int(''.join(filter(str.isdigit, 'Counters: 55')))
Out[2]: 55
In [3]: int(''.join(filter(str.isdigit, 'more than 23 times')))
Out[3]: 23
But be carefull !!! :
In [4]: int(''.join(filter(str.isdigit, '200 grams 5')))
Out[4]: 2005
I am amazed to see that no one has yet mentioned the usage of itertools.groupby as an alternative to achieve this.
You may use itertools.groupby() along with str.isdigit() in order to extract numbers from string as:
from itertools import groupby
my_str = "hello 12 hi 89"
l = [int(''.join(i)) for is_digit, i in groupby(my_str, str.isdigit) if is_digit]
The value hold by l will be:
[12, 89]
PS: This is just for illustration purpose to show that as an alternative we could also use groupby to achieve this. But this is not a recommended solution. If you want to achieve this, you should be using accepted answer of fmark based on using list comprehension with str.isdigit as filter.
Since none of these dealt with real world financial numbers in excel and word docs that I needed to find, here is my variation. It handles ints, floats, negative numbers, currency numbers (because it doesn’t reply on split), and has the option to drop the decimal part and just return ints, or return everything.
It also handles Indian Laks number system where commas appear irregularly, not every 3 numbers apart.
It does not handle scientific notation or negative numbers put inside parentheses in budgets — will appear positive.
It also does not extract dates. There are better ways for finding dates in strings.
import re
def find_numbers(string, ints=True):
numexp = re.compile(r'[-]?d[d,]*[.]?[d{2}]*') #optional - in front
numbers = numexp.findall(string)
numbers = [x.replace(',','') for x in numbers]
if ints is True:
return [int(x.replace(',','').split('.')[0]) for x in numbers]
else:
return numbers
# extract numbers from garbage string:
s = '12//n,_@#$%3.14kjlw0xdadfackvj1.6e-19&*ghn334'
newstr = ''.join((ch if ch in '0123456789.-e' else ' ') for ch in s)
listOfNumbers = [float(i) for i in newstr.split()]
print(listOfNumbers)
[12.0, 3.14, 0.0, 1.6e-19, 334.0]
I was looking for a solution to remove strings’ masks, specifically from Brazilian phones numbers, this post not answered but inspired me. This is my solution:
>>> phone_number = '+55(11)8715-9877'
>>> ''.join([n for n in phone_number if n.isdigit()])
'551187159877'
Using Regex below is the way
lines = "hello 12 hi 89"
import re
output = []
#repl_str = re.compile('d+.?d*')
repl_str = re.compile('^d+$')
#t = r'd+.?d*'
line = lines.split()
for word in line:
match = re.search(repl_str, word)
if match:
output.append(float(match.group()))
print (output)
with findall
re.findall(r'd+', "hello 12 hi 89")
['12', '89']
re.findall(r'bd+b', "hello 12 hi 89 33F AC 777")
['12', '89', '777']
I am just adding this answer because no one added one using Exception handling and because this also works for floats
a = []
line = "abcd 1234 efgh 56.78 ij"
for word in line.split():
try:
a.append(float(word))
except ValueError:
pass
print(a)
Output :
[1234.0, 56.78]
line2 = "hello 12 hi 89" # this is the given string
temp1 = re.findall(r'd+', line2) # find number of digits through regular expression
res2 = list(map(int, temp1))
print(res2)
you can search all the integers in the string through digit by using findall expression.
In the second step create a list res2 and add the digits found in string to this list.
To catch different patterns it is helpful to query with different patterns.
Setup all the patterns that catch different number patterns of interest:
- To find commas, e.g. 12,300 or 12,300.00
r'[d]+[.,d]+'
- To find floats, e.g. 0.123 or .123
r'[d]*[.][d]+'
- To find integers, e.g. 123
r'[d]+'
Combine with pipe ( | ) into one pattern with multiple or conditionals.
(Note: Put complex patterns first else simple patterns will return chunks of the complex catch instead of the complex catch returning the full catch).
p = '[d]+[.,d]+|[d]*[.][d]+|[d]+'
Below, we’ll confirm a pattern is present with re.search(), then return an iterable list of catches. Finally, we’ll print each catch using bracket notation to subselect the match object return value from the match object.
s = 'he33llo 42 I'm a 32 string 30 444.4 12,001'
if re.search(p, s) is not None:
for catch in re.finditer(p, s):
print(catch[0]) # catch is a match object
Returns:
33
42
32
30
444.4
12,001
For phone numbers you can simply exclude all non-digit characters with D in regex:
import re
phone_number = "(619) 459-3635"
phone_number = re.sub(r"D", "", phone_number)
print(phone_number)
The r in r"D" stands for raw string. It is necessary. Without it, Python will consider D as an escape character.
The cleanest way i found:
>>> data = 'hs122 125 &55,58, 25'
>>> new_data = ''.join((ch if ch in '0123456789.-e' else ' ') for ch in data)
>>> numbers = [i for i in new_data.split()]
>>> print(numbers)
['122', '125', '55', '58', '25']
or this:
>>> import re
>>> data = 'hs122 125 &55,58, 25'
>>> numbers = re.findall(r'd+', data)
>>> print(numbers)
['122', '125', '55', '58', '25']
str1 = "There are 2 apples for 4 persons"
# printing original string
print("The original string : " + str1) # The original string : There are 2 apples for 4 persons
# using List comprehension + isdigit() +split()
# getting numbers from string
res = [int(i) for i in str1.split() if i.isdigit()]
print("The numbers list is : " + str(res)) # The numbers list is : [2, 4]
I would like to extract all the numbers contained in a string. Which is better suited for the purpose, regular expressions or the isdigit() method?
Example:
line = "hello 12 hi 89"
Result:
[12, 89]
I’d use a regexp :
>>> import re
>>> re.findall(r'd+', "hello 42 I'm a 32 string 30")
['42', '32', '30']
This would also match 42 from bla42bla. If you only want numbers delimited by word boundaries (space, period, comma), you can use b :
>>> re.findall(r'bd+b', "he33llo 42 I'm a 32 string 30")
['42', '32', '30']
To end up with a list of numbers instead of a list of strings:
>>> [int(s) for s in re.findall(r'bd+b', "he33llo 42 I'm a 32 string 30")]
[42, 32, 30]
NOTE: this does not work for negative integers
I’m assuming you want floats not just integers so I’d do something like this:
l = []
for t in s.split():
try:
l.append(float(t))
except ValueError:
pass
Note that some of the other solutions posted here don’t work with negative numbers:
>>> re.findall(r'bd+b', 'he33llo 42 I'm a 32 string -30')
['42', '32', '30']
>>> '-3'.isdigit()
False
If you only want to extract only positive integers, try the following:
>>> txt = "h3110 23 cat 444.4 rabbit 11 2 dog"
>>> [int(s) for s in txt.split() if s.isdigit()]
[23, 11, 2]
I would argue that this is better than the regex example because you don’t need another module and it’s more readable because you don’t need to parse (and learn) the regex mini-language.
This will not recognize floats, negative integers, or integers in hexadecimal format. If you can’t accept these limitations, jmnas’s answer below will do the trick.
@jmnas, I liked your answer, but it didn’t find floats. I’m working on a script to parse code going to a CNC mill and needed to find both X and Y dimensions that can be integers or floats, so I adapted your code to the following. This finds int, float with positive and negative vals. Still doesn’t find hex formatted values but you could add “x” and “A” through “F” to the num_char tuple and I think it would parse things like ‘0x23AC’.
s = 'hello X42 I'm a Y-32.35 string Z30'
xy = ("X", "Y")
num_char = (".", "+", "-")
l = []
tokens = s.split()
for token in tokens:
if token.startswith(xy):
num = ""
for char in token:
# print(char)
if char.isdigit() or (char in num_char):
num = num + char
try:
l.append(float(num))
except ValueError:
pass
print(l)
This is more than a bit late, but you can extend the regex expression to account for scientific notation too.
import re
# Format is [(<string>, <expected output>), ...]
ss = [("apple-12.34 ba33na fanc-14.23e-2yapple+45e5+67.56E+3",
['-12.34', '33', '-14.23e-2', '+45e5', '+67.56E+3']),
('hello X42 I'm a Y-32.35 string Z30',
['42', '-32.35', '30']),
('he33llo 42 I'm a 32 string -30',
['33', '42', '32', '-30']),
('h3110 23 cat 444.4 rabbit 11 2 dog',
['3110', '23', '444.4', '11', '2']),
('hello 12 hi 89',
['12', '89']),
('4',
['4']),
('I like 74,600 commas not,500',
['74,600', '500']),
('I like bad math 1+2=.001',
['1', '+2', '.001'])]
for s, r in ss:
rr = re.findall("[-+]?[.]?[d]+(?:,ddd)*[.]?d*(?:[eE][-+]?d+)?", s)
if rr == r:
print('GOOD')
else:
print('WRONG', rr, 'should be', r)
Gives all good!
Additionally, you can look at the AWS Glue built-in regex
The best option I found is below. It will extract a number and can eliminate any type of char.
def extract_nbr(input_str):
if input_str is None or input_str == '':
return 0
out_number = ''
for ele in input_str:
if ele.isdigit():
out_number += ele
return float(out_number)
This answer also contains the case when the number is float in the string
def get_first_nbr_from_str(input_str):
'''
:param input_str: strings that contains digit and words
:return: the number extracted from the input_str
demo:
'ab324.23.123xyz': 324.23
'.5abc44': 0.5
'''
if not input_str and not isinstance(input_str, str):
return 0
out_number = ''
for ele in input_str:
if (ele == '.' and '.' not in out_number) or ele.isdigit():
out_number += ele
elif out_number:
break
return float(out_number)
If you know it will be only one number in the string, i.e 'hello 12 hi', you can try filter.
For example:
In [1]: int(''.join(filter(str.isdigit, '200 grams')))
Out[1]: 200
In [2]: int(''.join(filter(str.isdigit, 'Counters: 55')))
Out[2]: 55
In [3]: int(''.join(filter(str.isdigit, 'more than 23 times')))
Out[3]: 23
But be carefull !!! :
In [4]: int(''.join(filter(str.isdigit, '200 grams 5')))
Out[4]: 2005
I am amazed to see that no one has yet mentioned the usage of itertools.groupby as an alternative to achieve this.
You may use itertools.groupby() along with str.isdigit() in order to extract numbers from string as:
from itertools import groupby
my_str = "hello 12 hi 89"
l = [int(''.join(i)) for is_digit, i in groupby(my_str, str.isdigit) if is_digit]
The value hold by l will be:
[12, 89]
PS: This is just for illustration purpose to show that as an alternative we could also use groupby to achieve this. But this is not a recommended solution. If you want to achieve this, you should be using accepted answer of fmark based on using list comprehension with str.isdigit as filter.
Since none of these dealt with real world financial numbers in excel and word docs that I needed to find, here is my variation. It handles ints, floats, negative numbers, currency numbers (because it doesn’t reply on split), and has the option to drop the decimal part and just return ints, or return everything.
It also handles Indian Laks number system where commas appear irregularly, not every 3 numbers apart.
It does not handle scientific notation or negative numbers put inside parentheses in budgets — will appear positive.
It also does not extract dates. There are better ways for finding dates in strings.
import re
def find_numbers(string, ints=True):
numexp = re.compile(r'[-]?d[d,]*[.]?[d{2}]*') #optional - in front
numbers = numexp.findall(string)
numbers = [x.replace(',','') for x in numbers]
if ints is True:
return [int(x.replace(',','').split('.')[0]) for x in numbers]
else:
return numbers
# extract numbers from garbage string:
s = '12//n,_@#$%3.14kjlw0xdadfackvj1.6e-19&*ghn334'
newstr = ''.join((ch if ch in '0123456789.-e' else ' ') for ch in s)
listOfNumbers = [float(i) for i in newstr.split()]
print(listOfNumbers)
[12.0, 3.14, 0.0, 1.6e-19, 334.0]
I was looking for a solution to remove strings’ masks, specifically from Brazilian phones numbers, this post not answered but inspired me. This is my solution:
>>> phone_number = '+55(11)8715-9877'
>>> ''.join([n for n in phone_number if n.isdigit()])
'551187159877'
Using Regex below is the way
lines = "hello 12 hi 89"
import re
output = []
#repl_str = re.compile('d+.?d*')
repl_str = re.compile('^d+$')
#t = r'd+.?d*'
line = lines.split()
for word in line:
match = re.search(repl_str, word)
if match:
output.append(float(match.group()))
print (output)
with findall
re.findall(r'd+', "hello 12 hi 89")
['12', '89']
re.findall(r'bd+b', "hello 12 hi 89 33F AC 777")
['12', '89', '777']
I am just adding this answer because no one added one using Exception handling and because this also works for floats
a = []
line = "abcd 1234 efgh 56.78 ij"
for word in line.split():
try:
a.append(float(word))
except ValueError:
pass
print(a)
Output :
[1234.0, 56.78]
line2 = "hello 12 hi 89" # this is the given string
temp1 = re.findall(r'd+', line2) # find number of digits through regular expression
res2 = list(map(int, temp1))
print(res2)
you can search all the integers in the string through digit by using findall expression.
In the second step create a list res2 and add the digits found in string to this list.
To catch different patterns it is helpful to query with different patterns.
Setup all the patterns that catch different number patterns of interest:
- To find commas, e.g. 12,300 or 12,300.00
r'[d]+[.,d]+'
- To find floats, e.g. 0.123 or .123
r'[d]*[.][d]+'
- To find integers, e.g. 123
r'[d]+'
Combine with pipe ( | ) into one pattern with multiple or conditionals.
(Note: Put complex patterns first else simple patterns will return chunks of the complex catch instead of the complex catch returning the full catch).
p = '[d]+[.,d]+|[d]*[.][d]+|[d]+'
Below, we’ll confirm a pattern is present with re.search(), then return an iterable list of catches. Finally, we’ll print each catch using bracket notation to subselect the match object return value from the match object.
s = 'he33llo 42 I'm a 32 string 30 444.4 12,001'
if re.search(p, s) is not None:
for catch in re.finditer(p, s):
print(catch[0]) # catch is a match object
Returns:
33
42
32
30
444.4
12,001
For phone numbers you can simply exclude all non-digit characters with D in regex:
import re
phone_number = "(619) 459-3635"
phone_number = re.sub(r"D", "", phone_number)
print(phone_number)
The r in r"D" stands for raw string. It is necessary. Without it, Python will consider D as an escape character.
The cleanest way i found:
>>> data = 'hs122 125 &55,58, 25'
>>> new_data = ''.join((ch if ch in '0123456789.-e' else ' ') for ch in data)
>>> numbers = [i for i in new_data.split()]
>>> print(numbers)
['122', '125', '55', '58', '25']
or this:
>>> import re
>>> data = 'hs122 125 &55,58, 25'
>>> numbers = re.findall(r'd+', data)
>>> print(numbers)
['122', '125', '55', '58', '25']
str1 = "There are 2 apples for 4 persons"
# printing original string
print("The original string : " + str1) # The original string : There are 2 apples for 4 persons
# using List comprehension + isdigit() +split()
# getting numbers from string
res = [int(i) for i in str1.split() if i.isdigit()]
print("The numbers list is : " + str(res)) # The numbers list is : [2, 4]