CODEWITHSUNDEEP
pythonlistdictionarysplit
ksmakkapawee
ksmakkapawee

Reputation: 275

Python: Split list in array

Just beginning with python and know enough to know I know nothing. I would like to find alternative ways of splitting a list into a list of dicts. Example list:

data = ['**adjective:**', 'nice', 'kind', 'fine',
        '**noun:**', 'benefit', 'profit', 'advantage', 'avail', 'welfare', 'use', 'weal', 
        '**adverb:**', 'well', 'nicely', 'fine', 'right', 'okay']

I would be able to get:

[{'**adjective**': ('nice', 'kind', 'fine'),
 '**noun**': ('benefit', 'profit', 'advantage', 'avail', 'welfare', 'use', 'weal'),
 '**adverb**': ('well', 'nicely', 'fine', 'right', 'okay')}]

Upvotes: 2

Views: 2133

Answers (5)

user506710
user506710

Reputation:

If you assume inner to be the list of words you could have this as the code

data = ['adjective:', 'nice', 'kind', 'fine', 'noun:', 'benefit', 'profit', 'advantage', 'avail', 'welfare', 'use', 'weal', 'adverb:', 'well', 'nicely', 'fine', 'right', 'okay']

dict = {}

for x in data:

    if x[-1] == ':' :

       start = x.rstrip(':')

       dict[start] = []

    else:

       dict[start].append(x)

print dict

This prints the following dictionary

{'adjective': ['nice', 'kind', 'fine'], 'noun': ['benefit', 'profit', 'advantage', 'avail', 'welfare', 'use', 'weal'], 'adverb': ['well', 'nicely', 'fine', 'right', 'okay']}

Upvotes: 0

Sven Marnach
Sven Marnach

Reputation: 601529

This might be as close at it gets to what you have asked:

d = collections.defaultdict(list)
for s in data:
    if s.endswith(":"):
        key = s[:-1]
    else:
        d[key].append(s)
print d
# defaultdict(<type 'list'>, 
#     {'adjective': ['nice', 'kind', 'fine'], 
#      'noun': ['benefit', 'profit', 'advantage', 'avail', 'welfare', 'use', 'weal'], 
#      'adverb': ['well', 'nicely', 'fine', 'right', 'okay']})

Edit: Just for fun an alternative two-liner inspired by the answer by SilentGhost:

g = (list(v) for k, v in itertools.groupby(data, lambda x: x.endswith(':')))
d = dict((k[-1].rstrip(":"), v) for k, v in itertools.izip(g, g))

Upvotes: 10

eyquem
eyquem

Reputation: 27575

And if there were keys without elements of a list after them ? , I thought. So I added 'nada:' in front, 'nothing:' in the middle, and ’oops:’ at the end of the list named data.

Then, in these conditions, the code 1 (in following) with groupy appears to give a completely false result, the code 2 with defaultdict give a result in which keys 'nada:' , 'nothing:' , and ’oops:’ are absent. They are also less fast than the simplest solution (code 3: Cameron, user506710)

I had an idea => codes 4 and 5. Results are OK and executions are faster.

from time import clock

data = ['nada:',    # <<<=============
    'adjective:',
    'nice', 'kind', 'fine',
    'noun:',
    'benefit', 'profit', 'advantage', 'avail', 'welfare', 'use', 'weal',
    'nothing:', # <<<=============
    'adverb:',
    'well', 'nicely', 'fine', 'right', 'okay',
    'oops:'     # <<<=============
    ]

#------------------------------------------------------------
from itertools import groupby

te = clock()
dic1 = {}
for i, j in groupby(data, key=lambda x: x.endswith(':')):
    if i:
        key = next(j).rstrip(':')
        continue
    dic1[key] = list(j)
print clock()-te,'    groupby'
print dic1,'\n'

#------------------------------------------------------------
from collections import defaultdict
te = clock()
dic2 = defaultdict(list)
for s in data:
    if s.endswith(":"):
        key = s[:-1]
    else:
        dic2[key].append(s)
print clock()-te,'   defaultdict'
print dic2,'\n\n==================='

#=============================================================
te = clock()
dic4 = {}
for x in data:
    if x[-1] == ':' :
        start = x.rstrip(':')
        dic4[start] = []
    else:
    dic4[start].append(x)
print clock() - te
print dic4,'\n'

#------------------------------------------------------------
te = clock()
dic3 = {}
der = len(data)
for i,y in enumerate(data[::-1]):
    if y[-1]==':':
        dic3[y[0:-1]] = data[len(data)-i:der]
        der = len(data)-i-1
print clock()-te
print dic3,'\n'

    #------------------------------------------------------------
te = clock()
dic5 = {}
der = len(data)
for i in xrange(der-1,-1,-1):
    if data[i][-1]==':':
        dic5[data[i][0:-1]] = data[i+1:der]
        der = i
print clock() - te
print dic5

print '\ndic3==dic4==dic5 is',dic3==dic4==dic5

Upvotes: 0

SilentGhost
SilentGhost

Reputation: 319551

>>> data = ['adjective:', 'nice', 'kind', 'fine', 'noun:', 'benefit', 'profit', 'advantage', 'avail', 'welfare', 'use', 'weal', 'adverb:', 'well', 'nicely', 'fine', 'right', 'okay']
>>> from itertools import groupby
>>> dic = {}
>>> for i, j in groupby(data, key=lambda x: x.endswith(':')):
    if i:
        key = next(j).rstrip(':')
        continue
    dic[key] = list(j)

>>> dic
{'adjective': ['nice', 'kind', 'fine'], 'noun': ['benefit', 'profit', 'advantage', 'avail', 'welfare', 'use', 'weal'], 'adverb': ['well', 'nicely', 'fine', 'right', 'okay']}

Upvotes: 5

Cameron
Cameron

Reputation: 56

The code below will give you a dictionary with one entry for each word with a colon after it.

data = ['adjective:', 'nice', 'kind', 'fine', 'noun:', 'benefit', 'profit', 'advantage', 'avail', 'welfare', 'use', 'weal', 'adverb:', 'well', 'nicely', 'fine', 'right', 'okay']
result = {}
key = None
for item in data:
 if item.endswith(":"):
  key = item[:-1]
  result[key] = []
  continue
 result[key].append(item)

Upvotes: 1

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