Question

sample collection = 
[['', 'test1.txt', 'test2.txt', 'test3.txt', 'test4.txt'], 
['apple', 1, 1, 1, 1], 
['banana', 1, 1, 0, 0], 
['lemon', 1, 0, 2, 1]]

在进行tf-idf计算之前，我想收集每一行的“点击数”，也就是不为零的次数。

我尝试了以下操作，但被卡住了：

def n_calc(frequency_matrix):
    non_zero_term_frequencies = []
    for list_ in sample_collection:
        for value in list_:
            if value != 0.0:
                non_zero_term_frequencies += 1
            else:
                pass

idf_calc(sample_collection)

所需的输出：

non_zero_term_frequencies: [4, 2, 3]

Answer 1

使用列表理解，这应该可以工作：

counts = [
    sum(1 if value else 0 for value in line[1:]) for line in collection[1:]
]

print(counts)
>>> [4, 2, 3]

Answer 2

可以这样实现：

non_zero_term_frequencies = []
for list_ in frequency_matrix[1:]:
    non_zero = len(list_[1:]) \
               - len([x for x in list_ if x == 0])
    non_zero_term_frequencies.append(non_zero)

如何遍历矩阵并计算零项值？

2 个答案: