按最高频率键对字典排序

这个怎么样:

from collections import Counter

# input data
data = [{'123': 'abc'}, {'123': 'def'}, {'125': '123'}, {'125': '123243'}, {'125': '12312'}]

# count occurrence of first key in each dict in list
# this produces a dictionary with the data's keys
# ('123', '125', etc) as keys and the counts as values
counts = Counter([list(d.keys())[0] for d in data])

# sort by descending frequency of keys
data_sorted = sorted(
    data,
    key=lambda item: counts[list(item.keys())[0]], # function to lookup freq from counts
    reverse=True # descending order
)

print(data_sorted)

这导致:

[{'125': '123'}, {'125': '123243'}, {'125': '12312'}, {'123': 'abc'}, {'123': 'def'}]

获取字典键频率的最简单方法可能是 use a collections.Counter . 其工作原理如下:

from collections import Counter

freq = Counter(k for d in data for k in d)

现在 freq 是一个字典,其中每个键都作为一个键,该键出现的次数就是该值。使用 sorted key

freq_list = sorted(freq, key=freq.get)

使用 .reverse() 要将列表切换到降序:

freq_list.reverse()

最后,使用 功能,带有可选 钥匙 index 您的方法 freq_list

result = sorted(data, key = lambda d: freq_list.index(list(d)[0]))

解释 参数 已排序 已排序 钥匙

例如,当提供 get() 方法,结果将根据关联字典的自然顺序进行排序 get 返回给定键的值。当提供 index() 方法,则项目排序顺序将与该其他列表中的项目顺序匹配。