thelist = [{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}]
How do I say:
If "red" is in thelist and time does not equal 2 for that element (that's we just got from the list):
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Is there a difference in speed between "wallacoloo" answer and the list comprehension/pythonic answers?TIMEX– TIMEX2010-01-13 00:28:43 +00:00Commented Jan 13, 2010 at 0:28
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6 Answers
Using any() to find out if there is an element satisfying the conditions:
>>> any(item['color'] == 'red' and item['time'] != 2 for item in thelist)
False
3 Comments
Ben
this is nice! Am I understanding it correctly that the parentheses of the any() are doubling as a generator, so any() evaluates the list comprehension item by item and will return True once it finds an item that satisfies the criteria?
sth
It's a "generator expression", available in reasonably reason version of python (since 2.4). One could also write
any([...]) to get a classical list comprehension. (see python.org/dev/peps/pep-0289 for more on generator expressions)
orokusaki
That's the kind of stuff that makes me love Python.
def colorRedAndTimeNotEqualTo2(thelist):
for i in thelist:
if i["color"] == "red" and i["time"] != 2:
return True
return False
print colorRedAndTimeNotEqualTo2([{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}])
for i in thelist iterates through thelist, assigning the current element to i and doing the rest of the code in the block (for each value of i)
Thanks for the catch, Benson.
2 Comments
Well, there's nothing as elegant as "find" but you can use a list comprehension:
matches = [x for x in thelist if x["color"] == "red" and x["time"] != 2]
if len(matches):
m = matches[0]
# do something with m
However, I find the [0] and len() tedious. I often use a for loop with an array slice, such as:
matches = [x for x in thelist if x["color"] == "red" and x["time"] != 2]
for m in matches[:1]:
# do something with m
1 Comment
jpsimons
I was assuming an "at most one" kind of find. If you want all matches, I'd defer to someone else's answer.
list = [{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}]
for i in list:
if i['color'] == 'red' && i['time'] != 2:
print i
2 Comments
Benson
Perhaps this is too nitpicky, but I recommend you avoid naming things "list", because you may at some point need access to the list constructor and it would be a shame to have overwritten it.
Ponkadoodle
Python uses "and" instead of "&&"
for val in thelist:
if val['color'] == 'red' and val['time'] != 2:
#do something here
But it doesn't look like that's the right data structure to use.
Comments
You can do most of the list manipulation in a list comprehension. Here's one that makes a list of times for all elements where the color is red. Then you can ask if 2 exists in those times.
thelist = [{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}]
reds = ( x['time'] == 2 for x in thelist if x['color'] == red )
if False in reds:
do_stuff()
You can condense that even further by eliminating the variable "reds" like this:
thelist = [{'color':'green', 'time':4}, {'color':'red','time':2},{'color':'blue','time':5}]
if False in ( x['time'] == 2 for x in thelist if x['color'] == red ):
do_stuff()
2 Comments
jspcal
this requires 2 searches instead of 1 (one to get the element, another to find False). this also needs more ram (the list comprehension could generate a huge result). iterating would be more efficient
Benson
I switched to generator comprehensions rather than list comprehensions, since (as jspcal said) the lists were unnecessary.
