列表元素的计数器

Question

这里是 Python 的新手。

我正在寻找一种创建列表（输出）的简单方法，其中 returns 另一个 objective 列表 (MyList) 的元素计数，同时保留索引（？）。

这是我想要得到的：

MyList = ["a", "b", "c", "c", "a", "c"]
Output = [ 2 ,  1 ,  3 ,  3 ,  2 ,  3 ]

我找到了类似问题的解决方案。计算列表中每个元素出现的次数。

In  : Counter(MyList)
Out : Counter({'a': 2, 'b': 1, 'c': 3})

然而，这 returns 一个不保留索引的 Counter 对象。

我假设给定计数器中的键我可以构建我想要的输出，但是我不确定如何继续。

额外信息，我在脚本中导入了 pandas，而 MyList 实际上是 pandas 数据框中的一列。

Answer 1

您可以使用 list.count method, which will count the amount of times each string takes place in MyList. You can generate a new list with the counts by using a list comprehension:

MyList = ["a", "b", "c", "c", "a", "c"]

[MyList.count(i) for i in MyList]
# [2, 1, 3, 3, 2, 3]

Answer 2

你只需要实现下面的一段代码

    c=Counter(MyList)
    lout=[c[i] for i in MyList]

现在列出 lout 是您想要的输出

Answer 3

您可以使用函数 itemgetter:

而不是另一个解决方案中的 listcomp

from collections import Counter
from operator import itemgetter

lst = ["a", "b", "c", "c", "a", "c"]

c = Counter(lst)
itemgetter(*lst)(c)
# (2, 1, 3, 3, 2, 3)

更新：正如@ALollz 在评论中提到的，这个解决方案似乎是最快的解决方案。如果 OP 需要列表而不是元组，则必须将结果转换为 list.

Answer 4

这是 hettinger 的经典片段之一:)

from collections import Counter, OrderedDict

class OrderedCounter(Counter, OrderedDict):
     'Counter that remembers the order elements are first seen'
     def __repr__(self):
         return '%s(%r)' % (self.__class__.__name__,
                            OrderedDict(self))
     def __reduce__(self):
         return self.__class__, (OrderedDict(self),)

x = ["a", "b", "c", "c", "a", "c"]
oc = OrderedCounter(x)
>>> oc
OrderedCounter(OrderedDict([('a', 2), ('b', 1), ('c', 3)]))
>>> oc['a']
2

Answer 5

一个 pandas 解决方案如下所示：

df = pd.DataFrame(data=["a", "b", "c", "c", "a", "c"], columns=['MyList'])
df['Count'] = df.groupby('MyList')['MyList'].transform(len)

编辑：如果这是您唯一想做的事情，则不应使用 pandas。我只是因为 pandas 标签才回答了这个问题。

性能取决于组数：

MyList = np.random.randint(1, 10, 10000).tolist()
df = pd.DataFrame(MyList)

%timeit [MyList.count(i) for i in MyList]
# 1.32 s ± 15.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit df.groupby(0)[0].transform(len)
# 3.89 ms ± 112 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

MyList = np.random.randint(1, 9000, 10000).tolist()
df = pd.DataFrame(MyList)

%timeit [MyList.count(i) for i in MyList]
# 1.36 s ± 11.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit df.groupby(0)[0].transform(len)
# 1.33 s ± 19.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Answer 6

使用 np.unique 创建值计数字典并映射值。这会很快，但不如计数器方法快：

import numpy as np

list(map(dict(zip(*np.unique(MyList, return_counts=True))).get, MyList))
#[2, 1, 3, 3, 2, 3]

中等大小列表的一些时间安排：

MyList = np.random.randint(1, 2000, 5000).tolist()

%timeit [MyList.count(i) for i in MyList]
#413 ms ± 165 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit list(map(dict(zip(*np.unique(MyList, return_counts=True))).get, MyList))
#1.89 ms ± 1.73 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit pd.DataFrame(MyList).groupby(MyList).transform(len)[0].tolist()
#2.18 s ± 12.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

c=Counter(MyList)
%timeit lout=[c[i] for i in MyList]
#679 µs ± 2.33 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

c = Counter(MyList)
%timeit list(itemgetter(*MyList)(c))
#503 µs ± 162 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)

更大的列表：

MyList = np.random.randint(1, 2000, 50000).tolist()

%timeit [MyList.count(i) for i in MyList]
#41.2 s ± 5.27 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit list(map(dict(zip(*np.unique(MyList, return_counts=True))).get, MyList))
#18 ms ± 56.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit pd.DataFrame(MyList).groupby(MyList).transform(len)[0].tolist()
#2.44 s ± 12.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

c=Counter(MyList)
%timeit lout=[c[i] for i in MyList]
#6.89 ms ± 22.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

c = Counter(MyList)
%timeit list(itemgetter(*MyList)(c))
#5.27 ms ± 10.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 7

请注意，@Gio 指出列表是 pandas 系列对象。在这种情况下，您可以将 Series 对象转换为列表：

import pandas as pd

l = ["a", "b", "c", "c", "a", "c"]
ds = pd.Series(l) 
l=ds.tolist()
[l.count(i) for i in ds] 
# [2, 1, 3, 3, 2, 3]

但是，一旦你有了系列，你就可以通过 value_counts 来计算元素了。

l = ["a", "b", "c", "c", "a", "c"]
s = pd.Series(l) #Series object
c=s.value_counts() #c is Series again
[c[i] for i in s] 
# [2, 1, 3, 3, 2, 3]

列表元素的计数器

List elements’ counter

python

counter

list

python-3.x

pandas