为什么调用函数有效但断言函数会导致断言错误？

Question

我写了一个工作正常的函数，但是当我为它写一个断言代码时，它给出了一个断言错误。这段代码的唯一问题是修复断言错误。

def frequency(x, y):
    '''Get the frequency (the number of occurrences) of an element in a sequence.
     : param sequence: the sequence in which the element must be counted
     : param element: the element whose frequency we want to obtain
     : return: the frequency of the element in the sequence'''
    a = list(x)
    print(a.count(y))

def test_frequency():
    # Tests
    assert frequency('texts', 'e') == 1
    assert frequency('texts', 'a') == 0
    assert frequency('texts', 's') == 1
    assert frequency('texts', 't') == 2
    # limit tests
    assert frequency('ttt', 't') == 3
    assert frequency('', 'x') == 0

    print('test_frequency: ok')

test_frequency()
frequency(x = input("Enter a word: "), y = input("Enter a letter(symbol): "))

Answer 1

这个小问题是函数打印它的值，而不是 returning 它们。

Python 有一些相当可疑的行为 returning None 每当函数结束时没有 return 语句导致此类不幸的错误。

解决方案是在 return a.count(y) 而不是 print(a.count(y)) 结束。

Answer 2

这是因为您的函数正在打印值而不是返回计数。


def frequency(x, y):
    a = list(x)
    return a.count(y)

def test_frequency():
    # Tests
    assert frequency('texts', 'e') == 1
    assert frequency('texts', 'a') == 0
    assert frequency('texts', 's') == 1
    assert frequency('texts', 't') == 2
    # limit tests
    assert frequency('ttt', 't') == 3
    assert frequency('', 'x') == 0
    
    print('test_frequency: ok')

test_frequency()
frequency(x = input("Enter a word: "), y = input("Enter a letter(symbol): "))

为什么调用函数有效但断言函数会导致断言错误？

Why does calling the function work but asserting the function results in an assertion error?

python

assert