Question

我试图找到分割决策树的最佳变量，它需要分组并计算某些值的出现次数。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中每个都有多少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],…} .

我也对更好的聚合或数据结构方式持开放态度。这是我试图做的事情

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等属于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})}

计数器可让您.most_common()获取所需的列表：

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)]

如果你真的需要0计数，你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，但这种方式会出现任何问题，如果zipped

中没有，则为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)]

Answer 2

您可以尝试这样的事情：

data=[('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')]


manual_dict={}

for i,j in enumerate(data):
    if j[0] not in manual_dict:
        manual_dict[j[0]]=[j[1]]
    else:
        manual_dict[j[0]].append(j[1])


final_dict={}

for ia,aj in manual_dict.items():
    final_dict[ia]={'None':aj.count('None'),'Basic':aj.count('Basic'),'Premium':aj.count('Premium')}

print(final_dict)

输出：

{'c': {'Premium': 0, 'None': 1, 'Basic': 2}, 'a': {'Premium': 0, 'None': 3, 'Basic': 0}, 'd': {'Premium': 0, 'None': 1, 'Basic': 1}, 'b': {'Premium': 3, 'None': 1, 'Basic': 1}, 'e': {'Premium': 0, 'None': 1, 'Basic': 2}}

将列表列表分组并计算唯一第一个元素的第二个元素的出现次数

2 个答案: