Python 列表中的计数频率

我有一份清单：

countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

我想在上面的列表中找到子列表的频率

我已尝试使用itertools：

freq = [len(list(group)) for x in countall for key, group in groupby(x)]

但是，我得到了错误的结果：

[1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1]

---

我的清单理解有什么问题

正如ForceBru所指出的，首先对列表进行排序，然后使用groupby：

from itertools import groupby
countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

freq = [(key, len(list(x))) for key, x in groupby(sorted(countall))]
print(freq)

输出：

[([0, 5], 1), ([1, 4], 5), ([2, 3], 10), ([3, 2], 10), ([4, 1], 5), ([5, 0], 1)]

您的代码有错误：

freq = [len(list(group)) for x in countall for key, group in groupby(x)]
                       ^paranthesis missing

然后在

countall

中对每个单独的列表进行分组，这是不需要的

for x in countall for key, group in groupby(x)

您可以在排序（countall）上直接

groupby

---

此外，正如@Bemmu所回答的，您可以使用collections.Counter。但这不支持

列表

，因此首先必须将数据转换为元组或字符串，然后使用ForceBru指出的

计数器
，首先对列表排序，然后使用groupby：

from itertools import groupby
countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

freq = [(key, len(list(x))) for key, x in groupby(sorted(countall))]
print(freq)

输出：

[([0, 5], 1), ([1, 4], 5), ([2, 3], 10), ([3, 2], 10), ([4, 1], 5), ([5, 0], 1)]

您的代码有错误：

freq = [len(list(group)) for x in countall for key, group in groupby(x)]
                       ^paranthesis missing

然后在
countall
中对每个单独的列表进行分组，这是不需要的

for x in countall for key, group in groupby(x)

您可以在排序（countall）上直接
groupby

此外，正如@Bemmu所回答的，您可以使用collections.Counter。但它不支持
列表
，因此首先必须将数据转换为tuple或string，然后使用
计数器
Groupby似乎可以处理彼此后面的序列。要使用它，您需要首先对列表进行排序。另一个选项是使用该类：

输出：

Counter({(3, 2): 10, (2, 3): 10, (1, 4): 5, (4, 1): 5, (5, 0): 1, (0, 5): 1})

Groupby似乎处理的是一个接一个的序列。要使用它，您需要首先对列表进行排序。另一个选项是使用该类：

输出：

Counter({(3, 2): 10, (2, 3): 10, (1, 4): 5, (4, 1): 5, (5, 0): 1, (0, 5): 1})

如注释中所述，如果使用groupby，则需要对其进行排序

代码：

import itertools as it
freq = {tuple(key): len(list(group)) for key, group in it.groupby(sorted(countall))}

countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

print(freq)

{(3, 2): 10, (1, 4): 5, (2, 3): 10, (5, 0): 1, (0, 5): 1, (4, 1): 5}

测试代码：

import itertools as it
freq = {tuple(key): len(list(group)) for key, group in it.groupby(sorted(countall))}

countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

print(freq)

{(3, 2): 10, (1, 4): 5, (2, 3): 10, (5, 0): 1, (0, 5): 1, (4, 1): 5}

结果：

import itertools as it
freq = {tuple(key): len(list(group)) for key, group in it.groupby(sorted(countall))}

countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

print(freq)

{(3, 2): 10, (1, 4): 5, (2, 3): 10, (5, 0): 1, (0, 5): 1, (4, 1): 5}

如注释中所述，如果使用groupby，则需要对其进行排序

代码：

import itertools as it
freq = {tuple(key): len(list(group)) for key, group in it.groupby(sorted(countall))}

countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

print(freq)

{(3, 2): 10, (1, 4): 5, (2, 3): 10, (5, 0): 1, (0, 5): 1, (4, 1): 5}

测试代码：

import itertools as it
freq = {tuple(key): len(list(group)) for key, group in it.groupby(sorted(countall))}

countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

print(freq)

{(3, 2): 10, (1, 4): 5, (2, 3): 10, (5, 0): 1, (0, 5): 1, (4, 1): 5}

结果：

import itertools as it
freq = {tuple(key): len(list(group)) for key, group in it.groupby(sorted(countall))}

countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

print(freq)

{(3, 2): 10, (1, 4): 5, (2, 3): 10, (5, 0): 1, (0, 5): 1, (4, 1): 5}

您显示的代码语法无效，缺少括号。如果需要频率，请查看
集合。计数器
groupby
查找连续重复，因此您应该在
groupby
-ing之前对列表进行排序。您显示的代码语法无效，缺少括号。如果需要频率，请查看集合。计数器
groupby
查找连续重复，因此您应该在分组之前对列表进行排序。
集合。计数器
集合。计数器