Merging Two Dictionaries Values If They have The Same Keys-CodePudding

I am making a list word counter and trying to merge two dictionaries that have a counter for each character, but I keep getting the wrong outputs. Here is what I attempted in an effort to figure out why. Everything seems to go well except for the last two keys in the dictionary.

counts = {"a": 1, "p": 2, "l": 1, "e": 1}
new_counts = {"h": 1, "e": 1, "l": 2, "o": 1}

counts.update(new_counts)

for letters in counts:
    if letters in counts and new_counts:
        counts[letters]  = 1
    else:
        counts[letters] = 1

print(counts)

What I need:

{"a": 1, "p": 2, "l": 3, "e": 2, "h": 1, "o": 1}

What I get:

{'a': 2, 'p': 3, 'l': 3, 'e': 2, 'h': 2, 'o': 2}

CodePudding user response：

If we are in Python structures, actually it is a Counter that is ideal here.

from collections import Counter

c1 = Counter({"a": 1, "p": 2, "l": 1, "e": 1})
c2 = Counter({"h": 1, "e": 1, "l": 2, "o": 1})
print(c1   c2)

CodePudding user response：

You can use a simple for-loop:

counts = {"a": 1, "p": 2, "l": 1, "e": 1}
new_counts = {"h": 1, "e": 1, "l": 2, "o": 1}

for k, v in new_counts.items():
    if k in counts:
        counts[k]  = v
    else:
        counts[k] = v

print(counts)  # => {'a': 1, 'p': 2, 'l': 3, 'e': 2, 'h': 1, 'o': 1}

This is actually the fastest method. I tested it with timeit against kosciej16's answer and his took 5.49 seconds, while mine took 0.73 seconds (for one million iterations). I also tested against wim's dictionary comprehension answer and that took 1.95 seconds, while mine took 0.85 seconds.

CodePudding user response：

This would be an ideal place to use collections.defaultdict. Creating a new defaultdict, we iterate over both dictionaries' items and add their values to the defaultdict.

>>> from collections import defaultdict
>>> d = defaultdict(int)
>>> counts = {"a": 1, "p": 2, "l": 1, "e": 1}
>>> new_counts = {"h": 1, "e": 1, "l": 2, "o": 1}
>>> for k, v in counts.items():
...   d[k]  = v
... 
>>> for k, v in new_counts.items():
...   d[k]  = v
... 
>>> d
defaultdict(<class 'int'>, {'a': 1, 'p': 2, 'l': 3, 'e': 2, 'h': 1, 'o': 1})

CodePudding user response：

Using collections.Counter, a dict subclass designed for this kind of thing, you can just add instances directly using the operator:

>>> from collections import Counter
>>> counts = {"a": 1, "p": 2, "l": 1, "e": 1}
>>> new_counts = {"h": 1, "e": 1, "l": 2, "o": 1}
>>> Counter(counts)   Counter(new_counts)
Counter({'a': 1, 'p': 2, 'l': 3, 'e': 2, 'h': 1, 'o': 1})

Using a dict comprehension:

>>> all_keys = counts.keys() | new_counts.keys()
>>> {k: counts.get(k, 0)   new_counts.get(k, 0) for k in all_keys}
{'p': 2, 'h': 1, 'o': 1, 'e': 2, 'l': 3, 'a': 1}