Python - 计算矩阵行长度的频率

给定一个矩阵，任务是编写一个Python程序来获取其行长度的计数频率。

Input : test_list = [[6, 3, 1], [8, 9], [2], [10, 12, 7], [4, 11]]

Output : {3: 2, 2: 2, 1: 1}

Explanation : 2 lists of length 3 are present, 2 lists of size 2 and 1 of 1 length is present.

Input : test_list = [[6, 3, 1], [8, 9], [10, 12, 7], [4, 11]]

Output : {3: 2, 2: 2}

Explanation : 2 lists of length 3 are present, 2 lists of size 2.

方法#1：使用字典+循环

在这我们检查每一行的长度，如果长度已经出现在记忆字典中，那么结果就会增加，或者如果出现新的大小，则元素被注册为新的。

Python3

# Python3 code to demonstrate working of
# Row lengths counts
# Using dictionary + loop
  
# initializing list
test_list = [[6, 3, 1], [8, 9], [2], 
             [10, 12, 7], [4, 11]]
  
# printing original list
print("The original list is : " + str(test_list))
  
res = dict()
for sub in test_list:
  
    # initializing incase of new length
    if len(sub) not in res:
        res[len(sub)] = 1
  
    # increment in case of length present
    else:
        res[len(sub)] += 1
  
# printing result
print("Row length frequencies : " + str(res))

输出：

The original list is : [[6, 3, 1], [8, 9], [2], [10, 12, 7], [4, 11]]
Row length frequencies : {3: 2, 2: 2, 1: 1}

方法#2：使用Counter() + map() + len()

其中，map() 和 len() 获取矩阵中每个子列表的长度，Counter 用于保持每个长度的频率。

蟒蛇3

# Python3 code to demonstrate working of
# Row lengths counts
# Using Counter() + map() + len()
from collections import Counter
  
# initializing list
test_list = [[6, 3, 1], [8, 9], [2],
             [10, 12, 7], [4, 11]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# Counter gets the frequencies of counts
# map and len gets lengths of sublist
res = dict(Counter(map(len, test_list)))
  
# printing result
print("Row length frequencies : " + str(res))

输出：

The original list is : [[6, 3, 1], [8, 9], [2], [10, 12, 7], [4, 11]]
Row length frequencies : {3: 2, 2: 2, 1: 1}