Regex in Python Loop

Question

Trying to scrape weather condition (index 9 in list v) and save the variable for later. Having difficulty writing the proper regex to store condition that is either 1 or 2 words.

Tested my regex code on regexr.com and it looks fine but doesn't work when run in IDLE.

v = ['\n\n7:53 AM\n\n\n\n\n',
 '\n\n\n\n\n\n48 \nF\n    \n\n\n\n\n\n\n',
 '\n\n\n\n\n\n45 \nF\n    \n\n\n\n\n\n\n',
 '\n\n\n\n\n\n89 \n%\n    \n\n\n\n\n\n\n',
 '\n\nSE\n\n\n\n\n',
 '\n\n\n\n\n\n5 \nmph\n    \n\n\n\n\n\n\n',
 '\n\n\n\n\n\n0 \nmph\n    \n\n\n\n\n\n\n',
 '\n\n\n\n\n\n30.11 \nin\n    \n\n\n\n\n\n\n',
 '\n\n\n\n\n\n0.0 \nin\n    \n\n\n\n\n\n\n',
 '\n\nMostly Cloudy\n\n\n\n\n']

for condition in str(v[9]):
        condition_search = re.findall('[A-Z]\w+', condition)
        if len(condition_search) > 1:
            condition = ' '
            condition = condition.join(condition_search)
        else:
            condition = str(condition_search)

print(condition)

actual results:

'[]'

desired results

'Mostly Cloudy'

if re.search(r'^[A-Z]\w*(?:\s+[A-Z]\w*)?$', v[9].strip()): condition = v[9].strip() — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Oct 7, 2019 at 20:15
There's no need for the loop. And you can use join() with a list that has only 1 element, so you don't need the test. — Barmar
– Barmar, Commented Oct 7, 2019 at 20:19

tevemadar · Accepted Answer · 2019-10-07 21:13:36Z

3

Regexps are nice, but I think you are looking for .strip():

text='\n\nMostly Cloudy\n\n\n\n\n'
print(text.strip())

Result:

Mostly Cloudy

and the surrounding whitespace is gone.
(Find docs on https://docs.python.org/3/library/stdtypes.html)

answered Oct 7, 2019 at 21:13

tevemadar

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Comments

Emma Marcier · Accepted Answer · 2019-10-07 20:26:14Z

2

Maybe, this would simply return that:

import re
v = ['\n\n7:53 AM\n\n\n\n\n',
     '\n\n\n\n\n\n48 \nF\n    \n\n\n\n\n\n\n',
     '\n\n\n\n\n\n45 \nF\n    \n\n\n\n\n\n\n',
     '\n\n\n\n\n\n89 \n%\n    \n\n\n\n\n\n\n',
     '\n\nSE\n\n\n\n\n',
     '\n\n\n\n\n\n5 \nmph\n    \n\n\n\n\n\n\n',
     '\n\n\n\n\n\n0 \nmph\n    \n\n\n\n\n\n\n',
     '\n\n\n\n\n\n30.11 \nin\n    \n\n\n\n\n\n\n',
     '\n\n\n\n\n\n0.0 \nin\n    \n\n\n\n\n\n\n',
     '\n\nMostly Cloudy\n\n\n\n\n']

condition_search = re.findall(r'[A-Z][A-Za-z\s]+[a-z]', v[9])

print(condition_search[0])

Output

Mostly Cloudy

If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.

RegEx Circuit

jex.im visualizes regular expressions:

edited Oct 7, 2019 at 20:26

answered Oct 7, 2019 at 20:19

Emma Marcier

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2 Comments

Barmar Over a year ago

He wants to return a string, not a list.

Amocat _ Over a year ago

This is perfect, thank you all! I think if i use pop on above I can quickly convert to string.

the23Effect · Accepted Answer · 2019-10-07 21:45:58Z

2

Since you are doing scraping of some weather data am assuming the data you get is standardized in some way.

Looking at the data you can tell that information you need is padded by lots of newline and space characters at the front and back (which you don't need). To remove them:

Simpler Non-regex solution:

# This removes the leading and trailing white-space characters in each line,
# which also includes space, newline, tabs, etc,.
processed_weather_data = [line.strip() for line in v]

# Lets say you need weather condition which is at the 9th index
print(processed_weather_data[9])

answered Oct 7, 2019 at 21:45

the23Effect

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