Python Function returns wrong value

Question

periodsList = [] 
su = '0:'
Su = []
sun = []
SUN = ''

I'm formating timetables by converting

extendedPeriods = ['0: 1200 - 1500',
    '0: 1800 - 2330',
    '2: 1200 - 1500',
    '2: 1800 - 2330',
    '3: 1200 - 1500',
    '3: 1800 - 2330',
    '4: 1200 - 1500',
    '4: 1800 - 2330',
    '5: 1200 - 1500',
    '5: 1800 - 2330',
    '6: 1200 - 1500',
    '6: 1800 - 2330']

into '1200 - 1500/1800 - 2330'

su is the day identifier
Su, sun store some values

SUN stores the converted timetable

for line in  extendedPeriods:
  if su in line:    
    Su.append(line)

  for item in Su:
    sun.append(item.replace(su, '', 1).strip())

  SUN = '/'.join([str(x) for x in sun])

Then I tried to write a function to apply my "converter" also to the other days..

def formatPeriods(id, store1, store2, periodsDay): 
  for line in extendedPeriods:
    if id in line:    
      store1.append(line)

  for item in store1:
    store2.append(item.replace(id, '', 1).strip())

  periodsDay = '/'.join([str(x) for x in store2])  
  return periodsDay

But the function returns 12 misformatted strings...

'1200 - 1500', '1200 - 1500/1200 - 1500/1800 - 2330',

You should give your variables different names, you are bound to confuse them with these names. — Dahaka
– Dahaka, Commented Jul 11, 2013 at 10:15
What was the input you provided exactly to get those malformed strings? — mfjones
– mfjones, Commented Jul 11, 2013 at 10:18
Using your formatPeriods function with the inputs from the top of your posts, it works just as expected. — tobias_k
– tobias_k, Commented Jul 11, 2013 at 10:27

Ashwini Chaudhary · Accepted Answer · 2013-07-11 10:37:38Z

2

You can use collections.OrderedDict here, if order doesn't matter then use collections.defaultdict

>>> from collections import OrderedDict
>>> dic = OrderedDict()
for item in extendedPeriods:
    k,v = item.split(': ')
    dic.setdefault(k,[]).append(v)
...     
>>> for k,v in dic.iteritems():
...     print "/".join(v)
...     
1200 - 1500/1800 - 2330
1200 - 1500/1800 - 2330
1200 - 1500/1800 - 2330
1200 - 1500/1800 - 2330
1200 - 1500/1800 - 2330
1200 - 1500/1800 - 2330

To access a particular day you can use:

>>> print "/".join(dic['0'])   #sunday
1200 - 1500/1800 - 2330
>>> print "/".join(dic['2'])   #tuesday
1200 - 1500/1800 - 2330

edited Jul 11, 2013 at 10:37

answered Jul 11, 2013 at 10:31

Ashwini Chaudhary

252k60 gold badges478 silver badges519 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Burhan Khalid · Accepted Answer · 2013-07-11 10:30:54Z

1

This is your general logic:

from collections import defaultdict

d = defaultdict(list)

for i in extended_periods:
    bits = i.split(':')
    d[i[0].strip()].append(i[1].strip())

for i,v in d.iteritems():
   print i,'/'.join(v)

The output is:

0 1200 - 1500/1800 - 2330
3 1200 - 1500/1800 - 2330
2 1200 - 1500/1800 - 2330
5 1200 - 1500/1800 - 2330
4 1200 - 1500/1800 - 2330
6 1200 - 1500/1800 - 2330

To make it function for a day, simply select d[0] (for Sunday, for example):

def schedule_per_day(day):

    d = defaultdict(list)

    for i in extended_periods:
        bits = i.split(':')
        d[i[0].strip()].append(i[1].strip())

    return '/'.join(d[day]) if d.get(day) else None

answered Jul 11, 2013 at 10:30

Burhan Khalid

175k20 gold badges254 silver badges291 bronze badges

Collectives™ on Stack Overflow

Python Function returns wrong value

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related