CODEWITHSUNDEEP
pythonregexpython-2.7
Pasha
Pasha

Reputation: 190

python need help to extract pattern

From the following list, i am trying to extract only digits(ints & floats) & versions nums (only separated by dots).

[u'3.1.1', u'3.2', u'3.1.2', u'3', u'3.3.0', u'3.3.1-1', u'3.2.2', u'latest']

Tried the following code. Its not taking out 3.3.1-1. Need help with regex. Also is there any fastest way to do it?

 def myfunc(self, img_list):
    ret = list()
    for i in img_list:
        try:
            if re.match("([\d.]+)", i):
                ret.append(i)
            elif float(i):
                ret.append(i)
        except Exception as e:
            display.vvv("Error: %s" % str(e))
            pass

    return ret

Upvotes: 1

Views: 80

Answers (4)

daemon24
daemon24

Reputation: 1448

Just replace all . in the string and check if remaining characters are digits.

def myfunc(self , img_list):
    ret = list()
    for i in img_list:
        if '.' in i and i.replace('.','').isdigit():
            ret.append(i)
    return ret

Upvotes: 0

Mahesh Karia
Mahesh Karia

Reputation: 2055

Another solution using comprehension: 

lst = [u'3.1.1', u'3.2', u'3.1.2', u'3', u'3.3.0', u'3.3.1-1', u'3.2.2', u'latest']
results = [i for i in lst if i.replace('.', '').isdigit()]
print results

output:

[u'3.1.1', u'3.2', u'3.1.2', u'3', u'3.3.0', u'3.2.2']

Upvotes: 1

whackamadoodle3000
whackamadoodle3000

Reputation: 6748

You could try a list comprehension.

lst = [u'3.1.1', u'3.2', u'3.1.2', u'3', u'3.3.0', u'3.3.1-1', u'3.2.2', u'latest']
print([e for e in lst if ''.join(e.split('.')).isdigit()])

Prints:

['3.1.1', '3.2', '3.1.2', '3', '3.3.0', '3.2.2']

Upvotes: 0

Xero Smith
Xero Smith

Reputation: 2076

Regex is one way to go. You can accomplish the same using filter

>>> lst = [u'3.1.1', u'3.2', u'3.1.2', u'3', u'3.3.0', u'3.3.1-1', u'3.2.2', u'latest']
>>> list(filter(lambda s: all(c.isdigit() or c == "." for c in s), lst))
['3.1.1', '3.2', '3.1.2', '3', '3.3.0', '3.2.2']

Upvotes: 0

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