CODEWITHSUNDEEP
pythonlistdictionary
eamon1234
eamon1234

Reputation: 1585

Splitting dictionary of lists based on criteria in python

I have data from travel diaries which has been read in from a csv file. I have it set up as a dictionary with a bunch of lists. E.g.:

print diary['ID'][1] gives 123456789
print diary['TravelReferenceDay'][1] gives 1 for a Monday

I want to randomnly select an ID from the array based on the day e.g.:

random.choice(diary['ID']) if diary['TravelReferenceDay'] == 1

I can arrange the data by TravelReferenceDay in the csv file. I had tried the groupby method to split up the array:

groups = []
uniquekeys = []
for k, g in groupby(diary, diary['TravelReferenceDay']):
    groups.append(list(g))      # Store group iterator as a list
    uniquekeys.append(k)

But that gave the error:

TypeError: 'list' object is not callable

Could you suggest a way to achieve this? Thanks.

Upvotes: 0

Views: 372

Answers (2)

KurzedMetal
KurzedMetal

Reputation: 12946

My solution with a list comprehensions:

In [1]: import random
   ...: diary = {'ID': ['11', '22', '33', '44', '55'], 'TravelReferenceDay': [1, 1, 2, 3, 1]}
   ...: monday_diary = [x for n, x in enumerate(diary['ID']) if diary['TravelReferenceDay'][n] == 1]

In [2]: monday_diary
Out[2]: ['11', '22', '55']

In [3]: random.choice(monday_diary)
Out[3]: '22'

Upvotes: 1

ecatmur
ecatmur

Reputation: 157344

The second argument to groupby is a callable that is invoked on successive items from the iterable first argument.

You want to use operator.itemgetter('TravelReferenceDay'):

for k, g in groupby(diary, operator.itemgetter('TravelReferenceDay')):
    ...

This is equivalent to lambda x: x['TravelReferenceDay'].

Note that groupby expects the iterable to already be sorted by the key; groups contain adjacent items with the same key.

OK, this won't work because you've stored your data as parallel arrays. For ease of processing I'd advise to convert it to a list of dicts:

diary = [dict((k, diary[k][i]) for k in diary) for i in range(len(diary['ID']))]

Upvotes: 2

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