Python代码实现列表分组计数

def count_by(arr, fn=lambda x: x):
  key = {}
  for el in map(fn, arr):
    key[el] = 1 if el not in key else key[el] + 1
  return key

# EXAMPLES
from math import floor
count_by([6.1, 4.2, 6.3], floor) # {6: 2, 4: 1}
count_by(['one', 'two', 'three'], len) # {3: 2, 5: 1}

{ key_expr: [x for x in collection2 if condition2] for value in collection1 if condition1 }

>>> d = {'one': 1, 'two': 2, 'three': 3, 'two': 2}
>>> d
{'one': 1, 'two': 2, 'three': 3}
>>> d = {'one': 1, 'two': 2, 'three': 3, 'two': 22}
>>> d
{'one': 1, 'two': 22, 'three': 3}
>>>

def count_by(lst, fn):
  return {key : len([el for el in lst if fn(el) == key]) for key in map(fn, lst)}

class collections.defaultdict([default_factory[, ...]])

 from collections import defaultdict

def count_by(lst, fn):
  d = defaultdict(int)
  for el in lst:
    d[fn(el)] += 1
  return d

def group_by(lst, fn):
  d = defaultdict(list)
  for el in lst:
    d[fn(el)].append(el)
  return d

# EXAMPLES
from math import floor
group_by([6.1, 4.2, 6.3], floor) # {4: [4.2], 6: [6.1, 6.3]}
group_by(['one', 'two', 'three'], len) # {3: ['one', 'two'], 5: ['three']}