Reputation: 3181
I have
{key1:value1, key2:value2, etc}
I want it to become:
[key1,value1,key2,value2] , if certain keys match certain criteria.
How can i do it as pythonically as possible?
Thanks!
Upvotes: 4
Views: 256
Reputation: 113935
This code should solve your problem:
myList = []
for tup in myDict.iteritems():
myList.extend(tup)
>>> myList
[1, 1, 2, 2, 3, 3]
Upvotes: 2
Reputation: 103814
If speed matters, use extend to add the key, value pairs to an empty list:
l=[]
for t in sorted(d.items()):
return l.extend(t)
>>> d={'key1':'val1','key2':'val2'}
>>> l=[]
>>> for t in sorted(d.items()):
... l.extend(t)
...
>>> l
['key1', 'val1', 'key2', 'val2']
Not only faster, this form is easier to add logic to each key, value pair.
Speed comparison:
d={'key1':'val1','key2':'val2'}
def f1():
l=[]
for t in d.items():
return l.extend(t)
def f2():
return [y for x in d.items() for y in x]
cmpthese.cmpthese([f1,f2])
Prints:
rate/sec f2 f1
f2 908,348 -- -33.1%
f1 1,358,105 49.5% --
Upvotes: 0
Reputation: 143047
Another entry/answer:
import itertools
dict = {'one': 1, 'two': 2}
bl = [[k, v] for k, v in dict.items()]
list(itertools.chain(*bl))
yields
['two', 2, 'one', 1]
Upvotes: 1
Reputation: 13347
given a dict
, this will combine all items to a tuple
sum(dict.items(),())
if you want a list rather than a tuple
list(sum(dict.items(),()))
for example
dict = {"We": "Love", "Your" : "Dict"}
x = list(sum(dict.items(),()))
x
is then
['We', 'Love', 'Your', 'Dict']
Upvotes: 6
Reputation: 142136
The most efficient (not necessarily most readable or Python is)
from itertools import chain
d = { 3: 2, 7: 9, 4: 5 } # etc...
mylist = list(chain.from_iterable(d.iteritems()))
Apart from materialising the lists, everything is kept as iterators.
Upvotes: 2
Reputation: 224904
This should do the trick:
[y for x in dict.items() for y in x]
For example:
dict = {'one': 1, 'two': 2}
print([y for x in dict.items() for y in x])
This will print:
['two', 2, 'one', 1]
Upvotes: 11
Reputation: 3985
>>> a = {"lol": 1 }
>>> l = []
>>> for k in a.keys():
... l.append( k )
... l.append( a[k] )
...
>>> l
['lol', 1]
Upvotes: 0