统计嵌套列表python中的数据

Question

我有一个嵌套的列表。在列表中，它有一个日期和一个名称。我想要日期的用户计数。例如，在'2019-10-11‘中，有5个用户。但是sam用户已经存在，所以我将其算作4。

a =[['sam', '2019-10-11'], ['ram', '2019-10-12'], ['king', '2019-10-11'],
    ['queen', '2019-10-10'], ['ram', '2019-10-12'], ['kumar', '2019-10-11'],
    ['jeba', '2019-10-11'], ['sam', '2019-10-10'], ['sam', '2019-10-11'],
    ['howl', '2019-10-14']]

需要的输出：

a = [['2019-10-10','2019-10-11','2019-10-12','2019-10-14'],[2,4,1,1]]

Answer 1

您可以使用下面给出的代码，它迭代数组并计算json b中的键。

a =[['sam', '2019-10-11'], ['ram', '2019-10-12'], ['king', '2019-10-11'],
    ['queen', '2019-10-10'], ['ram', '2019-10-12'], ['kumar', '2019-10-11'],
    ['jeba', '2019-10-11'], ['sam', '2019-10-10'], ['sam', '2019-10-11'],
    ['howl', '2019-10-14']]

b = {}

for iter in a:
  if iter[1] in b:
    if(iter[0] not in b[iter[1]]):
      b[iter[1]][iter[0]] = 1
  else:
    b[iter[1]] = {iter[0]: 1}

    print(b)

上述代码的结果将是：

{'2019-10-11': {'sam': 1, 'king': 1, 'kumar': 1, 'jeba': 1}, '2019-10-12': {'ram': 1}, '2019-10-10': {'queen': 1, 'sam': 1}, '2019-10-14': {'howl': 1}}

当然，您可以将其转换为如下所示的数组结构

for key in b:
  print(key, len(b[key]))
  res[0].append(key)
  res[1].append(len(b[key]))

其结果将是

[['2019-10-11', '2019-10-12', '2019-10-10', '2019-10-14'], [4, 1, 2, 1]]

Answer 2

你可以使用collections中的defaultdict，比如

>>> a =[['sam', '2019-10-11'], ['ram', '2019-10-12'], ['king', '2019-10-11'],
...     ['queen', '2019-10-10'], ['ram', '2019-10-12'], ['kumar', '2019-10-11'],
...     ['jeba', '2019-10-11'], ['sam', '2019-10-10'], ['sam', '2019-10-11'],
...     ['howl', '2019-10-14']]
>>> 
>>> from collections import defaultdict
>>> d = defaultdict(set) # using `set` to not care about duplicate `name` on same `date`, otherwise use `list` instead of `set`
>>> # and use `d[date].append(name)`
>>> for name, date in a:
...   d[date].add(name) # maybe, two users can be there for the same date ?
... 
>>> data = [(k,len(v)) for k,v in sorted(d.items())]
>>> data
[('2019-10-10', 2), ('2019-10-11', 4), ('2019-10-12', 1), ('2019-10-14', 1)]
>>> list(zip(*data)) # returns a `list` of `tuple`s which usually should be fine for any iteration
[('2019-10-10', '2019-10-11', '2019-10-12', '2019-10-14'), (2, 4, 1, 1)]
>>> required = [list(x) for x in zip(*data)] # if you really want `list` of `list`s
>>> required
[['2019-10-10', '2019-10-11', '2019-10-12', '2019-10-14'], [2, 4, 1, 1)]]

Answer 3

from collections import Counter

list(zip(*Counter(list(zip(*set(zip(*zip(*a)))))[1]).items()))

Out[54]: [('2019-10-11', '2019-10-10', '2019-10-12', '2019-10-14'), (4, 2, 1, 1)]

这可以简化为：

list(zip(*Counter([j for _, j in set([tuple(i) for i in a])]).items()))

[('2019-10-11', '2019-10-10', '2019-10-12', '2019-10-14'), (4, 2, 1, 1)]

不导入模块的另一种方法：

b = set(zip(*zip(*a)))
d = {}

for name,date in b:
    d[date] = d.get(date,0)+1
list(zip(*d.items()))

[('2019-10-11', '2019-10-10', '2019-10-12', '2019-10-14'), (4, 2, 1, 1)]

如果你需要列表的排序列表，你可以这样做：

[list(i) for i in zip(*sorted(d.items(),key=lambda x:x[0]))]
Out[112]: [['2019-10-10', '2019-10-11', '2019-10-12', '2019-10-14'], [2, 4, 1, 1]]