One trivial solution is to choose a method dependent on an estimate of how completely values are covered by dictionary keys. General case Use df[‘A’].map(d) if all values mapped; or Use df[‘A’].map(d).fillna(df[‘A’]).astype(int) if >5% values mapped. Few, e.g. < 5%, values in d Use df[‘A’].replace(d) The “crossover point” of ~5% is specific to Benchmarking below. … Read more
You can use tools like pylint to check whether your code is PEP8 compatible or not : so I tried something like : M_D1 = {” foo bar “:1, ” a bc”:2, ” “:3} and pylint gave me 10/10 rating, so I guess it has no issues with the spaces used in keys.
The solution is relatively simple: x = Dict(“a”=>”A”, “b”=>”B”, “c”=>”C”) for (key, value) in x print(key); print(value) end # Output: cCbBaA Check out the Julia docs for Base.Dict to learn more about functions you can apply to a Dictionary in Julia!
Are your ranges non-overlapping? If so you could use a TreeMap: TreeMap<Double, Character> m = new TreeMap<Double, Character>(); m.put(1.0, ‘A’); m.put(2.9, null); m.put(4.0, ‘B’); m.put(6.0, null); m.put(6.5, ‘C’); m.put(10.0, null); The lookup logic is a bit complicated by the fact that you probably want an inclusive lookup (i.e. 2.9 maps to ‘A’, and not undefined): … Read more
The difference is that the second method does not work: >>> {}.update(1, 2) Traceback (most recent call last): File “<stdin>”, line 1, in <module> TypeError: update expected at most 1 arguments, got 2 dict.update() expects to find a iterable of key-value pairs, keyword arguments, or another dictionary: Update the dictionary with the key/value pairs from … Read more
You might want to try a few of your scenarios using evaluate, it might do what you are looking for. def stringMap = “[‘a’:2,’b’:4]” def map = evaluate(stringMap) assert map.a == 2 assert map.b == 4 def stringMapNested = “[‘foo’:’bar’, baz:[‘alpha’:’beta’]]” def map2 = evaluate(stringMapNested) assert map2.foo == “bar” assert map2.baz.alpha == “beta”
If you don’t want to use an immutable library you can create a new Map on change (this is a shallow copy of the map, like spreading an object): const [someMap, setSomeMap] = useState(new Map()) And when you need to update it: setSomeMap(new Map(someMap.set(‘someKey’, ‘a new value’))) The same concept applies to Redux: case ‘SomeAction’: … Read more
A little benchmark for you (ipython): In [1]: def test_1(d, k): …: if k in d: …: var1 = d[k] …: In [2]: def test_2(d, k): …: if d.has_key(k): …: var1 = d[k] …: In [3]: def test_3(d, k): …: try: …: var1 = d[k] …: except KeyError as e: …: pass …: In [4]: … Read more
Simply do: _objTable.removeWhere((key, value) => key == “propertyName”); It works with nested Maps too.
Short answer: because they are mutable containers. If a dict was hashed, its hash would change as you changed its contents.