Suppose I had a string
string1 = "498results should get"
Now I need to get only integer values from the string like 498. Here I don't want to use list slicing because the integer values may increase like these examples:
string2 = "49867results should get"
string3 = "497543results should get"
So I want to get only integer values out from the string exactly in the same order. I mean like 498,49867,497543 from string1,string2,string3 respectively.
Can anyone let me know how to do this in a one or two lines?
>>> import re
>>> string1 = "498results should get"
>>> int(re.search(r'\d+', string1).group())
498
If there are multiple integers in the string:
>>> map(int, re.findall(r'\d+', string1))
[498]
An answer taken from ChristopheD here: https://stackoverflow.com/a/2500023/1225603
r = "456results string789"
s = ''.join(x for x in r if x.isdigit())
print int(s)
456789
Here's your one-liner, without using any regular expressions, which can get expensive at times:
>>> ''.join(filter(str.isdigit, "1234GAgade5312djdl0"))
returns:
'123453120'
if you have multiple sets of numbers then this is another option
>>> import re
>>> print(re.findall('\d+', 'xyz123abc456def789'))
['123', '456', '789']
its no good for floating point number strings though.
Iterator version
>>> import re
>>> string1 = "498results should get"
>>> [int(x.group()) for x in re.finditer(r'\d+', string1)]
[498]
>>> import itertools
>>> int(''.join(itertools.takewhile(lambda s: s.isdigit(), string1)))
With python 3.6, these two lines return a list (may be empty)
>>[int(x) for x in re.findall('\d+', your_string)]
Similar to
>>list(map(int, re.findall('\d+', your_string))
this approach uses list comprehension, just pass the string as argument to the function and it will return a list of integers in that string.
def getIntegers(string):
numbers = [int(x) for x in string.split() if x.isnumeric()]
return numbers
Like this
print(getIntegers('this text contains some numbers like 3 5 and 7'))
Output
[3, 5, 7]
def function(string):
final = ''
for i in string:
try:
final += str(int(i))
except ValueError:
return int(final)
print(function("4983results should get"))
Another option is to remove the trailing the letters using rstrip and string.ascii_lowercase (to get the letters):
import string
out = [int(s.replace(' ','').rstrip(string.ascii_lowercase)) for s in strings]
Output:
[498, 49867, 497543]
integerstring=""
string1 = "498results should get"
for i in string1:
if i.isdigit()==True
integerstring=integerstring+i
print(integerstring)
Related
I'm trying to separate numbers from a string in Python such as this example:
text = "Compute the average of 5,7". (I want to get a list [5,7])
(the comma between the numbers is a must) I've tried using:
numbers = [int(i) for i in text.split() if i.isdigit()]
It works when the numbers aren't separated by the comma but when written with a comma I just receive an empty list.
Use a regular expression to find two integers separated by comma.
import re
m = re.search(r'(\d+),(\d+)', text)
if m:
numbers = [int(x) for x in m.groups()]
Try this:
text = "Compute the average of 5,7"
nums = [int(i) for i in text if i.isdigit()]
print(nums)
# prints [5, 7]
>>> import re
>>> re.search('.*(\d,\d).*', text).group(1)
'5,7'
How do I concatenate a list of strings into a single string?
For example, given ['this', 'is', 'a', 'sentence'], how do I get "this-is-a-sentence"?
For handling a few strings in separate variables, see How do I append one string to another in Python?.
For the opposite process - creating a list from a string - see How do I split a string into a list of characters? or How do I split a string into a list of words? as appropriate.
Use str.join:
>>> words = ['this', 'is', 'a', 'sentence']
>>> '-'.join(words)
'this-is-a-sentence'
>>> ' '.join(words)
'this is a sentence'
A more generic way (covering also lists of numbers) to convert a list to a string would be:
>>> my_lst = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> my_lst_str = ''.join(map(str, my_lst))
>>> print(my_lst_str)
12345678910
It's very useful for beginners to know
why join is a string method.
It's very strange at the beginning, but very useful after this.
The result of join is always a string, but the object to be joined can be of many types (generators, list, tuples, etc).
.join is faster because it allocates memory only once. Better than classical concatenation (see, extended explanation).
Once you learn it, it's very comfortable and you can do tricks like this to add parentheses.
>>> ",".join("12345").join(("(",")"))
Out:
'(1,2,3,4,5)'
>>> list = ["(",")"]
>>> ",".join("12345").join(list)
Out:
'(1,2,3,4,5)'
Edit from the future: Please don't use the answer below. This function was removed in Python 3 and Python 2 is dead. Even if you are still using Python 2 you should write Python 3 ready code to make the inevitable upgrade easier.
Although #Burhan Khalid's answer is good, I think it's more understandable like this:
from str import join
sentence = ['this','is','a','sentence']
join(sentence, "-")
The second argument to join() is optional and defaults to " ".
list_abc = ['aaa', 'bbb', 'ccc']
string = ''.join(list_abc)
print(string)
>>> aaabbbccc
string = ','.join(list_abc)
print(string)
>>> aaa,bbb,ccc
string = '-'.join(list_abc)
print(string)
>>> aaa-bbb-ccc
string = '\n'.join(list_abc)
print(string)
>>> aaa
>>> bbb
>>> ccc
We can also use Python's reduce function:
from functools import reduce
sentence = ['this','is','a','sentence']
out_str = str(reduce(lambda x,y: x+"-"+y, sentence))
print(out_str)
We can specify how we join the string. Instead of '-', we can use ' ':
sentence = ['this','is','a','sentence']
s=(" ".join(sentence))
print(s)
If you have a mixed content list and want to stringify it, here is one way:
Consider this list:
>>> aa
[None, 10, 'hello']
Convert it to string:
>>> st = ', '.join(map(str, map(lambda x: f'"{x}"' if isinstance(x, str) else x, aa)))
>>> st = '[' + st + ']'
>>> st
'[None, 10, "hello"]'
If required, convert back to the list:
>>> ast.literal_eval(st)
[None, 10, 'hello']
If you want to generate a string of strings separated by commas in final result, you can use something like this:
sentence = ['this','is','a','sentence']
sentences_strings = "'" + "','".join(sentence) + "'"
print (sentences_strings) # you will get "'this','is','a','sentence'"
def eggs(someParameter):
del spam[3]
someParameter.insert(3, ' and cats.')
spam = ['apples', 'bananas', 'tofu', 'cats']
eggs(spam)
spam =(','.join(spam))
print(spam)
Without .join() method you can use this method:
my_list=["this","is","a","sentence"]
concenated_string=""
for string in range(len(my_list)):
if string == len(my_list)-1:
concenated_string+=my_list[string]
else:
concenated_string+=f'{my_list[string]}-'
print([concenated_string])
>>> ['this-is-a-sentence']
So, range based for loop in this example , when the python reach the last word of your list, it should'nt add "-" to your concenated_string. If its not last word of your string always append "-" string to your concenated_string variable.
I have a list that contains counts for different items as strings. Sometimes a restock is given in brackets. The list looks like this:
23
21 (+3)
32 (+14)
So there is always a space between the number and the brackets. To start of I've wrote a function that I apply over the Series. I've used the split-method two times to return only the first number:
splitted = item.split(" ")
splitted2 = splitted[0].split("+")
return int(splitted2[0])
This solution is kind of hacky already and on top of that, I am missing the restocks in brackets. Now I want to know, how I would possibly add both of the numbers together. I would therefore split the list one time so that I get this as a result:
['23']
['21', '(+3)']
Then I want to select only the list items, that have to values, get rid of the + and () and add the first and the second value together. How would I do that?
With help of this question Extract numbers from a string
str = "32 (+14)"
import re
data = re.findall(r'\d+', str)
# ['32', '14']
sum = 0
for d in data:
sum+=int(d)
print(sum)
Output:
46
See how numbers are parsed using regex. Here, d represents [0-9]
Another method: (with help of this answer)
>>> import re
>>> str = "32 (+14)"
>>> eval(re.sub('[\(\)]', '', str.replace(' ', '')))
46
This will also allow user to do any operation
This is one way, which literally evaluates the '+' operation.
import ast, re
lst = ['23', '21 (+3)', '32 (+14)']
lst = [ast.literal_eval(re.sub('[\(\)]', '', i.replace(' ', ''))) for i in lst]
# [23, 24, 46]
not the best method, but this works. regex will be better option.
>>> str = "32 (+14)"
>>> str.replace("(+", "").replace(")", "").split(" ")
['32', '14']
Regex:
>>> import re
>>> str = "32 (+14)"
>>> nums = re.findall('\d+', str )
>>> print(sum(int(i) for i in nums))
46
How do I concatenate a list of strings into a single string?
For example, given ['this', 'is', 'a', 'sentence'], how do I get "this-is-a-sentence"?
For handling a few strings in separate variables, see How do I append one string to another in Python?.
For the opposite process - creating a list from a string - see How do I split a string into a list of characters? or How do I split a string into a list of words? as appropriate.
Use str.join:
>>> words = ['this', 'is', 'a', 'sentence']
>>> '-'.join(words)
'this-is-a-sentence'
>>> ' '.join(words)
'this is a sentence'
A more generic way (covering also lists of numbers) to convert a list to a string would be:
>>> my_lst = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> my_lst_str = ''.join(map(str, my_lst))
>>> print(my_lst_str)
12345678910
It's very useful for beginners to know
why join is a string method.
It's very strange at the beginning, but very useful after this.
The result of join is always a string, but the object to be joined can be of many types (generators, list, tuples, etc).
.join is faster because it allocates memory only once. Better than classical concatenation (see, extended explanation).
Once you learn it, it's very comfortable and you can do tricks like this to add parentheses.
>>> ",".join("12345").join(("(",")"))
Out:
'(1,2,3,4,5)'
>>> list = ["(",")"]
>>> ",".join("12345").join(list)
Out:
'(1,2,3,4,5)'
Edit from the future: Please don't use the answer below. This function was removed in Python 3 and Python 2 is dead. Even if you are still using Python 2 you should write Python 3 ready code to make the inevitable upgrade easier.
Although #Burhan Khalid's answer is good, I think it's more understandable like this:
from str import join
sentence = ['this','is','a','sentence']
join(sentence, "-")
The second argument to join() is optional and defaults to " ".
list_abc = ['aaa', 'bbb', 'ccc']
string = ''.join(list_abc)
print(string)
>>> aaabbbccc
string = ','.join(list_abc)
print(string)
>>> aaa,bbb,ccc
string = '-'.join(list_abc)
print(string)
>>> aaa-bbb-ccc
string = '\n'.join(list_abc)
print(string)
>>> aaa
>>> bbb
>>> ccc
We can also use Python's reduce function:
from functools import reduce
sentence = ['this','is','a','sentence']
out_str = str(reduce(lambda x,y: x+"-"+y, sentence))
print(out_str)
We can specify how we join the string. Instead of '-', we can use ' ':
sentence = ['this','is','a','sentence']
s=(" ".join(sentence))
print(s)
If you have a mixed content list and want to stringify it, here is one way:
Consider this list:
>>> aa
[None, 10, 'hello']
Convert it to string:
>>> st = ', '.join(map(str, map(lambda x: f'"{x}"' if isinstance(x, str) else x, aa)))
>>> st = '[' + st + ']'
>>> st
'[None, 10, "hello"]'
If required, convert back to the list:
>>> ast.literal_eval(st)
[None, 10, 'hello']
If you want to generate a string of strings separated by commas in final result, you can use something like this:
sentence = ['this','is','a','sentence']
sentences_strings = "'" + "','".join(sentence) + "'"
print (sentences_strings) # you will get "'this','is','a','sentence'"
def eggs(someParameter):
del spam[3]
someParameter.insert(3, ' and cats.')
spam = ['apples', 'bananas', 'tofu', 'cats']
eggs(spam)
spam =(','.join(spam))
print(spam)
Without .join() method you can use this method:
my_list=["this","is","a","sentence"]
concenated_string=""
for string in range(len(my_list)):
if string == len(my_list)-1:
concenated_string+=my_list[string]
else:
concenated_string+=f'{my_list[string]}-'
print([concenated_string])
>>> ['this-is-a-sentence']
So, range based for loop in this example , when the python reach the last word of your list, it should'nt add "-" to your concenated_string. If its not last word of your string always append "-" string to your concenated_string variable.
i have a list:
a = ['a','b','c'.........'A','B','C'.........'Z']
and i have string:
string1= 's#$%ERGdfhliisgdfjkskjdfW$JWLI3590823r'
i want to keep ONLY those characters in string1 that exist in a
what is the most effecient way to do this? perhaps instead of having a be a list, i should just make it a string? like this a='abcdefg..........ABC..Z' ??
This should be faster.
>>> import re
>>> string1 = 's#$%ERGdfhliisgdfjkskjdfW$JWLI3590823r'
>>> a = ['E', 'i', 'W']
>>> r = re.compile('[^%s]+' % ''.join(a))
>>> print r.sub('', string1)
EiiWW
This is even faster than that.
>>> all_else = ''.join( chr(i) for i in range(256) if chr(i) not in set(a) )
>>> string1.translate(None, all_else)
'EiiWW'
44 microsec vs 13 microsec on my laptop.
How about that?
(Edit: turned out, translate yields the best performance.)
''.join([s for s in string1 if s in a])
Explanation:
[s for s in string1 if s in a]
creates a list of all characters in string1, but only if they are also in the list a.
''.join([...])
turns it back into a string by joining it with nothing ('') in between the elements of the given list.
List comprehension to the rescue!
wanted = ''.join(letter for letter in string1 if letter in a)
(Note that when passing a list comprehension to a function you can omit the brackets so that the full list isn't generated prior to being evaluated. While semantically the same as a list comprehension, this is called a generator expression.)
If, you are going to do this with large strings, there is a faster solution using translate; see this answer.
#katrielalex: To spell it out:
import string
string1= 's#$%ERGdfhliisgdfjkskjdfW$JWLI3590823r'
non_letters= ''.join(chr(i) for i in range(256) if chr(i) not in string.letters)
print string1.translate(None,non_letters)
print 'Simpler, but possibly less correct'
print string1.translate(None, string.punctuation+string.digits+string.whitespace)