The usual way is to create a new dictionary containing only the items you want to keep: new_data = {k: v for k, v in data.items() if v[0] <= 30} If you need to change the original dictionary in place, you can use a for-loop: for k, v in list(data.items()): if v[0] > 30: del … Read more
You can use tuple unpacking in for statement: for i, (key, value) in enumerate(a.iteritems()): # Do something with i, key, value >>> d = {‘a’: ‘b’} >>> for i, (key, value) in enumerate(d.iteritems()): … print i, key, value … 0 a b Side Note: In Python 3.x, use dict.items() which returns an iterable dictionary view. … Read more
Casting from int to an enum is extremely cheap… it’ll be faster than a dictionary lookup. Basically it’s a no-op, just copying the bits into a location with a different notional type. Parsing a string into an enum value will be somewhat slower. I doubt that this is going to be a bottleneck for you … Read more
Try adding .to_dict(‘records’)[0] to the row you want catStr_df[catStr_df[‘Item Number’]==34881].to_dict(‘records’)[0]
HashMap<String, String> x; Random random = new Random(); List<String> keys = new ArrayList<String>(x.keySet()); String randomKey = keys.get( random.nextInt(keys.size()) ); String value = x.get(randomKey);
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.
>>> from collections import defaultdict >>> l= [(1,4),(2,4),(3,4),(4,15),(5,15),(6,23),(7,23),(8,23),(9,15),(10,23),(11,15),(12,15)] >>> d= defaultdict( list ) >>> for v, k in l: … d[k].append(v) … >>> d defaultdict(<type ‘list’>, {23: [6, 7, 8, 10], 4: [1, 2, 3], 15: [4, 5, 9, 11, 12]}) >>> [ {k:d[k]} for k in sorted(d) ] [{4: [1, 2, 3]}, {15: [4, … Read more
From the python documentation: A dictionary’s keys are almost arbitrary values. Values that are not hashable, that is, values containing lists, dictionaries or other mutable types (that are compared by value rather than by object identity) may not be used as keys. Hashable is defined as follows An object is hashable if it has a … Read more
To solve this you should use Python’s defaultdict. The first time you use a key that doesn’t exist, the argument to the defaultdict constructor is used to create a value (in this case, a list). http://docs.python.org/library/collections.html#defaultdict-examples from collections import defaultdict d = defaultdict(list) for i in range( 0, 10 ): for j in range( 0, … Read more