Question

如果我有dict个列表，例如：

{
    'id1': ['a', 'b', 'c'],
    'id2': ['a', 'b'],
    # etc.
}

我希望计算列表的大小，即ids的数量＆gt; 0，＆gt; 1，＆gt; 2 ...等等

是否有比嵌套for循环更简单的方法：

dictOfOutputs = {}
for x in range(1,11):
    count = 0
    for agentId in userIdDict:
        if len(userIdDict[agentId]) > x:
            count += 1
    dictOfOutputs[x] = count        
return dictOfOutputs

Answer 1

我使用collections.Counter() object来收集长度，然后积累总和：

from collections import Counter

lengths = Counter(len(v) for v in userIdDict.values())
total = 0
accumulated = {}
for length in range(max(lengths), -1, -1):
    count = lengths.get(length, 0)
    total += count
    accumulated[length] = total

因此，这会收集每个长度的计数，然后构建一个累积长度的字典。这是一个O（N）算法;你遍历所有的值一次，然后添加一些较小的直线循环（对于max()和累积循环）：

>>> from collections import Counter
>>> import random
>>> testdata = {''.join(random.choice('abcdefghijklmnopqrstuvwxyz') for _ in range(5)): [None] * random.randint(1, 10) for _ in range(100)}
>>> lengths = Counter(len(v) for v in testdata.values())
>>> lengths
Counter({8: 14, 7: 13, 2: 11, 3: 10, 4: 9, 5: 9, 9: 9, 10: 9, 1: 8, 6: 8})
>>> total = 0
>>> accumulated = {}
>>> for length in range(max(lengths), -1, -1):
...     count = lengths.get(length, 0)
...     total += count
...     accumulated[length] = total
... 
>>> accumulated
{0: 100, 1: 100, 2: 92, 3: 81, 4: 71, 5: 62, 6: 53, 7: 45, 8: 32, 9: 18, 10: 9}

Answer 2

是的，还有更好的方法。

首先，按照数据的长度索引ID：

my_dict = {
    'id1': ['a', 'b', 'c'],
    'id2': ['a', 'b'],
}

from collections import defaultdict
ids_by_data_len = defaultdict(list)

for id, data in my_dict.items():
    my_dict[len(data)].append(id)

现在，创建你的词典：

output_dict = {}
accumulator = 0
# note: the end of a range is non-inclusive!
for data_len in reversed(range(1, max(ids_by_data_len.keys()) + 1):
    accumulator += len(ids_by_data_len.get(data_len, []))
    output_dict[data_len-1] = accumulator

这具有O（n）复杂度而不是O（n²），因此对于大型数据集来说它也快得多。

如何用dict计算列表的大小？

2 个答案: