列表理解 - 尝试从键值对数组的数组中创建字典

Question

While trying to create a dict using list comprehension (without overwriting generated list keys)

x = {}
entries = [[1,'1a'], [2,'2a'], [3,'3a'], ['1', '1b'], ['2', '2b'], ['3', '3b']]
discardable = [x.setdefault(entry[0], []).append(entry[1]) for entry in entries]

Error: name 'x' is not defined

I was expecting x to be populated as:

{1: ['1a', '1b'], 2: ['2a', '2b'], 3: ['3a', '3b']}

How to explain this / make this work? Is there any other way to achieve this?

Answer 1

You can try:

from collections import defaultdict
entries = [['1','1a'], [2,'2a'], [3,'3a'], ['1', '1b'], ['2', '2b'], ['3', '3b']]
x = defaultdict(list)
[x[a].append(b) for a,b in entries]
x = dict(x)

---

Output

{'1': ['1a', '1b'], 2: ['2a'], 3: ['3a'], '2': ['2b'], '3': ['3b']}

Answer 2

Some of the keys are str and some are int , so this will produce

{'1': ['1a', '1b'], 2: ['2a'], 3: ['3a'], '2': ['2b'], '3': ['3b']}

You need to cast entry[0] to int

x = {}
entries = [[1,'1a'], [2,'2a'], [3,'3a'], ['1', '1b'], ['2', '2b'], ['3', '3b']]
[x.setdefault(int(entry[0]), []).append(entry[1]) for entry in entries]

print(x) will give

{1: ['1a', '1b'], 2: ['2a', '2b'], 3: ['3a', '3b']}

Answer 3

What you want is to merge the value of (1, '1'), (2, '2'), (3, '3'). The code you provided works, but it doesn't handle the type between int and str . All you need to do is cast str into int so the value would be merged.

Try this:

x = {}
entries = [[1,'1a'], [2,'2a'], [3,'3a'], ['1', '1b'], ['2', '2b'], ['3', '3b']]
discardable = [x.setdefault(int(entry[0]), []).append(entry[1]) for entry in entries]

Answer 4

How to explain this / make this work?

A comprehension creates one output for each input, and they're not intended for mutation

Is there any other way to achieve this?

Use a normal procedural loop. Or use a multidict but the standard library doesn't provide one.

You could perform functional transformations until you get the proper "shape" (using sorted , itertools.groupby and some more mapping) but I don't think that'd be worth it, a bog-standard for loop would be much more readable:

for k, v in entries:
    x.setdefault(int(x), []).append(v)

versus:

x = {
    x: list(v[1] for v in vs)
    for x, vs in itertools.groupby(
        sorted((int(k), v) for k, v in entries),
        lambda it: it[0]
    )
}

the comprehension is less readable and less efficient

Answer 5

You can use a dict comprehension . they work much like list comprehensions

entries = [[1,'1a'], [2,'2a'], [3,'3a'], ['1', '1b'], ['2', '2b'], ['3', '3b']]

# Build a dictionary from the values in entries
discardable = {key:val for key, val in entries}

print (discardable)
# Result: {1: '1a', 2: '2a', 3: '3a', '1': '1b', '2': '2b', '3': '3b'}