You can do this: s = dict([ (k,r) for k,r in mydict.iteritems() if r[‘x’] > 92 and r[‘x’] < 95 and r[‘y’] > 70 and r[‘y’] < 75 ]) This takes a dict as you specified and returns a ‘filtered’ dict.
Let’s take a look at your list comprehension then, but first let’s start with list comprehension at it’s easiest. l = [1,2,3,4,5] print [x for x in l] # prints [1, 2, 3, 4, 5] You can look at this the same as a for loop structured like so: for x in l: print x … Read more
The JSON might contain a list of objects, which needs to be searched: Python 2.7 version: def item_generator(json_input, lookup_key): if isinstance(json_input, dict): for k, v in json_input.iteritems(): if k == lookup_key: yield v else: for child_val in item_generator(v, lookup_key): yield child_val elif isinstance(json_input, list): for item in json_input: for item_val in item_generator(item, lookup_key): yield item_val … Read more
You can use set and set comprehension: {hobby for name, hobbies in input.items() for hobby in hobbies} As m.wasowski mentioned, we don’t use the name here, so we can use item.values() instead: {hobby for hobbies in input.values() for hobby in hobbies} If you really need a list as the result, you can do this (but … Read more
You must be looping over the wrong data set; just loop directly over the JSON-loaded dictionary, there is no need to call .keys() first: data = json.loads(response) myList = [item for item in data if item == “number1″] You may want to use u”number1” to avoid implicit conversions between Unicode and byte strings: data = … Read more
The method list.sort() is sorting the list in place, and as all mutating methods it returns None. Use the built-in function sorted() to return a new sorted list. result = sorted((trans for trans in my_list if trans.type in types), key=lambda x: x.code) Instead of lambda x: x.code, you could also use the slightly faster operator.attrgetter(“code”).
Use zip() and sum(): In [1]: l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)] In [2]: [sum(x) for x in zip(*l)] Out[2]: [25, 20] or: In [4]: map(sum, zip(*l)) Out[4]: [25, 20] timeit results: In [16]: l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]*1000 In [17]: %timeit … Read more
List comprehensions are more compact and faster than an explicit for loop building a list: def slower(): result = [] for elem in some_iterable: result.append(elem) return result def faster(): return [elem for elem in some_iterable] This is because calling .append() on a list causes the list object to grow (in chunks) to make space for … Read more
You can use tuple for doing that job like this: [(print(“bla1”), print(“bla2”)) for i in list] it’s work correctly.