Python – 因子频率字典

给定一个包含元素的列表，构造一个具有因子频率的字典。

Input : test_list = [2, 4, 6, 8]
Output : {1: 4, 2: 4, 3: 1, 4: 2, 5: 0, 6: 1, 7: 0, 8: 1}
Explanation : All factors count mapped, e.g 2 is divisible by all 4 values, hence mapped with 4.

Input : test_list = [1, 2]
Output : {1: 2, 2 : 1}
Explanation : Similar as above, 1 is factor of all.

方法#1：使用循环

这是可以执行此任务的粗暴方式。在此，元素被迭代并检查所需的数量是否是一个因素，如果是，则其在字典中对应于其键的频率增加。

Python3

# Python3 code to demonstrate working of 
# Factors Frequency Dictionary
# Using loop
  
# initializing list
test_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]
  
# printing original list
print("The original list : " + str(test_list))
  
res = dict()
  
# iterating till max element 
for idx in range(1, max(test_list)):
    res[idx] = 0
    for key in test_list:
          
        # checking for factor 
        if key % idx == 0:
            res[idx] += 1
          
# printing result 
print("The constructed dictionary : " + str(res))

输出

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]
The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 7: 0, 8: 2, 9: 2, 10: 0, 11: 0, 12: 1, 13: 0, 14: 0, 15: 1, 16: 1, 17: 0}

方法 #2：使用 sum() + 循环

这与上述问题的方法几乎相似。不同之处在于 sum() 用于求和，而不是用于解决问题的手动循环。

Python3

# Python3 code to demonstrate working of 
# Factors Frequency Dictionary
# Using sum() + loop
  
# initializing list
test_list = [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]
  
# printing original list
print("The original list : " + str(test_list))
  
res = dict()
for idx in range(1, max(test_list)):
      
    # using sum() instead of loop for sum computation
    res[idx] = sum(key % idx == 0 for key in test_list)
  
# printing result 
print("The constructed dictionary : " + str(res))

输出

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]
The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 7: 0, 8: 2, 9: 2, 10: 0, 11: 0, 12: 1, 13: 0, 14: 0, 15: 1, 16: 1, 17: 0}