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):
views:
114answers:
6
+1
A:
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.
Wallacoloo
2010-01-13 00:22:31
beaten to the punch by 30 seconds-- this is exactly what I would do. Except, just to be in the python 3 habit: print(i)
Ben
2010-01-13 00:24:01
That's workable, but the use of a for loop means that if there was more than one element with color == red and time != 2 you'd get I printed twice.
Benson
2010-01-13 00:28:55
A:
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()
Benson
2010-01-13 00:24:42
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
jspcal
2010-01-13 21:53:12
I switched to generator comprehensions rather than list comprehensions, since (as jspcal said) the lists were unnecessary.
Benson
2010-01-14 23:22:54
A:
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
darkporter
2010-01-13 00:25:41
I was assuming an "at most one" kind of find. If you want all matches, I'd defer to someone else's answer.
darkporter
2010-01-13 00:27:15
A:
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
jspcal
2010-01-13 00:25:59
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.
Benson
2010-01-13 21:20:55
Wallacoloo
2010-03-25 22:26:56
A:
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.
job
2010-01-13 00:27:17
+11
A:
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
sth
2010-01-13 00:33:42
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?
Ben
2010-01-13 00:37:17
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 http://www.python.org/dev/peps/pep-0289/ for more on generator expressions)
sth
2010-01-13 00:42:13