python - 对列表的列表进行分组，并计算唯一的第一个元素的第二个元素的出现次数

Question

我试图找到分割决策树的最佳变量，它需要对某些值的出现进行分组和计数。 虚拟数据集是

zipped=[(‘a’, ‘None’), (‘b’, ‘Premium’), (‘c’, ‘Basic’), (‘d’, ‘Basic’), (‘b’, ‘Premium’), (‘e’, ‘None’), (‘e’, ‘Basic’), (‘b’, ‘Premium’), (‘a’, ‘None’), (‘c’, ‘None’), (‘b’, ‘None’), (‘d’, ‘None’), (‘c’, ‘Basic’), (‘a’, ‘None’), (‘b’, ‘Basic’), (‘e’, ‘Basic’)]

所以，我想知道 a、b、c、d、e 中分别有多少个 None、Basic 和 Premium 我需要它看起来像

{‘a’:[‘None’:3,‘Basic’:0,‘Premium’:0], ‘b’:[‘None’:1,‘Basic’:1,‘Premium’:3],…} .

我也愿意接受更好的聚合或数据结构方式。 这是我尝试做的事情

temp=Counter( x[1] for x in zipped  if x[0]=='b')
print(temp)

我得到了

Counter({'Premium': 3, 'None': 1, 'Basic': 1})

Answer 1

假设您的 a、b 等是您的 slashdot、google:

zipped=[('a', 'None'), ('b', 'Premium'), ('c', 'Basic'), ('d', 'Basic'), ('b', 'Premium'), 
        ('e', 'None'), ('e', 'Basic'), ('b', 'Premium'), ('a', 'None'), ('c', 'None'), 
        ('b', 'None'), ('d', 'None'), ('c', 'Basic'), ('a', 'None'), ('b', 'Basic'), 
        ('e', 'Basic')]


from collections import Counter

d = {}
for key,val in zipped:
    d.setdefault(key,[]).append(val) # create key with empty list (if needed) + append val.

# now they are ordered lists, overwrite with Counter of it:
for key in d:
    d[key] = Counter(d[key])

print(d)

输出:

 {'a': Counter({'None': 3}), 
  'b': Counter({'Premium': 3, 'None': 1, 'Basic': 1}), 
  'c': Counter({'Basic': 2, 'None': 1}), 
  'd': Counter({'Basic': 1, 'None': 1}), 
  'e': Counter({'Basic': 2, 'None': 1})}

Counter 为您提供 .most_common() 来获取您想要的列表:

for k in d:
    print(k,d[k].most_common()) 

输出:

a [('None', 3)]
b [('Premium', 3), ('None', 1), ('Basic', 1)]
c [('Basic', 2), ('None', 1)]
d [('Basic', 1), ('None', 1)]
e [('Basic', 2), ('None', 1)]

如果您确实需要 0 计数，您可以在事后添加它们:

allVals = {v for _,v in zipped}   # get distinct values of zipped
for key in d:
    for v in allVals:
        d[key].update([v])        # add value once
        d[key].subtract([v])      # subtract value once

有点麻烦，但这样所有的东西都会出现，如果 zipped 中不存在则值为 0

for k in d:
    print(k,d[k].most_common()) 

输出:

a [('None', 3), ('Premium', 0), ('Basic', 0)]
b [('Premium', 3), ('None', 1), ('Basic', 1)]
c [('Basic', 2), ('None', 1), ('Premium', 0)]
d [('Basic', 1), ('None', 1), ('Premium', 0)]
e [('Basic', 2), ('None', 1), ('Premium', 0)]