You generally use .get with a default get(key[, default]) Return the value for key if key is in the dictionary, else default. If default is not given, it defaults to None, so that this method never raises a KeyError. So when you use get the loop would look like this: for item in newS: functionField … Read more
You can use indent option in json.dumps() to obtain \n symbols: import json user_dict = {‘name’: ‘dinesh’, ‘code’: ‘dr-01’} user_encode_data = json.dumps(user_dict, indent=2).encode(‘utf-8′) print(user_encode_data) Output: b'{\n “name”: “dinesh”,\n “code”: “dr-01″\n}’
The default implementation will compare the addresses stored by the pointers, so different objects will be considered as different keys. However, the logical state of the object will not be considered. For example, if you use std::string * as the key, two different std::string objects with the same text of “Hello” would be considered a … Read more
You didn’t say how exactly you want to merge, so take your pick: x = {‘both1’: 1, ‘both2’: 2, ‘only_x’: 100} y = {‘both1’: 10, ‘both2’: 20, ‘only_y’: 200} print {k: x.get(k, 0) + y.get(k, 0) for k in set(x)} print {k: x.get(k, 0) + y.get(k, 0) for k in set(x) & set(y)} print {k: … Read more
This is actually quite tricky – particularly if you want a useful error message when things are inconsistent, while correctly accepting duplicate but consistent entries (something no other answer here does..) Assuming you don’t have huge numbers of entries, a recursive function is easiest: def merge(a: dict, b: dict, path=[]): for key in b: if … Read more
To answer your question in one sentence: Per default, Maps don’t have a last entry, it’s not part of their contract. And a side note: it’s good practice to code against interfaces, not the implementation classes (see Effective Java by Joshua Bloch, Chapter 8, Item 52: Refer to objects by their interfaces). So your declaration … Read more
Cong Ma does a good job of explaining what __getitem__ is used for – but I want to give you an example which might be useful. Imagine a class which models a building. Within the data for the building it includes a number of attributes, including descriptions of the companies that occupy each floor : … Read more
>>> sorted(cityPopulation.iteritems(),key=lambda (k,v): v[0],reverse=True) #1990 [(‘C’, [30, 10, 20]), (‘B’, [20, 30, 10]), (‘A’, [10, 20, 30])] >>> sorted(cityPopulation.iteritems(),key=lambda (k,v): v[2],reverse=True) #2010 [(‘A’, [10, 20, 30]), (‘C’, [30, 10, 20]), (‘B’, [20, 30, 10])] Note in python 3 you can’t automagically unpack lambda arguments so you would have to change the code sorted(cityPopulation.items(), key=lambda k_v: … Read more
.ToDictionary(kvp=>kvp.Key,kvp=>kvp.Value); Isn’t that much more work.
Here’s another solution using a custom action, if you want to specify dict key pairs together comma-separated — import argparse import sys parser = argparse.ArgumentParser(description=’parse key pairs into a dictionary’) class StoreDictKeyPair(argparse.Action): def __call__(self, parser, namespace, values, option_string=None): my_dict = {} for kv in values.split(“,”): k,v = kv.split(“=”) my_dict[k] = v setattr(namespace, self.dest, my_dict) parser.add_argument(“–key_pairs”, … Read more