封装和泛化函数 def mult_tasks(str_numbers) in python

Question

问题：

在不改变 mult_tasks 的情况下，为 mult_tasks_line 写一个定义，这样 doctests 通过。
所以：

print mult_tasks("3469") 将产生：

(3*3) (3*4) (3*6) (3*9)
(4*3) (4*4) (4*6) (4*9)
(6*3) (6*4) (6*6) (6*9)
(9*3) (9*4) (9*6) (9*8)

def mult_tasks_line(first_num, str_numbers):

    """
    >>> mult_tasks_line("4", "3469")
    '(4*3) (4*4) (4*6) (4*9) '
    """
    #Add your code here

def mult_tasks(str_numbers):

    """
    >>> mult_tasks("246")
    '(2*2) (2*4) (2*6) \n(4*2) (4*4) (4*6) \n(6*2) (6*4) (6*6) \n'
    >>> mult_tasks("1234")
    '(1*1) (1*2) (1*3) (1*4) \n(2*1) (2*2) (2*3) (2*4) \n(3*1) (3*2) (3*3) (3*4) \n(4*1) (4*2) (4*3) (4*4) \n'
    """

    #Do not alter any code in this function
    result_tasks = ""
    for ch in str_numbers:
        result_tasks += mult_tasks_line(ch, str_numbers) + "\n"
    return result_tasks


if __name__ == "__main__":
    import doctest
    doctest.testmod(verbose = True)

我已经尝试这样做了：

def mult_tasks_line(first_num, str_numbers):
    digits=0
    for num in str_numbers:
        while digits<len(str_numbers):
            print "("+first_num+"*"+str_numbers[digits]+")",
            digits=digits+1

def mult_tasks(str_numbers):
    result_tasks = ""
    for ch in str_numbers:
        result_tasks += mult_tasks_line(ch, str_numbers),"\n"
    return result_tasks

这是我试过的，第一个函数工作得非常接近，但它没有单引号。当运行 mult_tasks_line("4", "3469") 结果 (4*3) (4*4) (4*6) (4*9)。

但是第二个功能似乎完全错误。这是第二个函数的结果：

mult_tasks("246”)

(2*2) (2*4) (2*6) 

Traceback (most recent call last):
File "<pyshell#1>", line 1, in <module> mult_tasks("246")
File "/Users/HuesFile/Downloads/Mastery Test/2.py", line 25, in mult_tasks
result_tasks += mult_tasks_line(ch, str_numbers),"\n" 
TypeError: cannot concatenate 'str' and 'tuple' objects

Answer 1

据我了解，您想从一串数字字符中打印有序元组的矩阵，而分隔符

def mult_tasks_line(first_num, str_numbers):
    return ''.join(
        ['({n1}*{n2}) '.format(n1 = first_num, n2 = n2 ) for n2 in str_numbers])

def mult_tasks(str_numbers):
    result_tasks = ""
    for ch in str_numbers: 
        result_tasks += mult_tasks_line(ch, str_numbers) + "\n" 
    return result_tasks

然后

print mult_tasks("3469")

给予

(3*3) (3*4) (3*6) (3*9) 
(4*3) (4*4) (4*6) (4*9) 
(6*3) (6*4) (6*6) (6*9) 
(9*3) (9*4) (9*6) (9*9)

随心所欲。从适用于任意字符串的意义上说，结果是可缩放的。

您希望计算乘法吗？

Answer 2

显然 mult_tasks 包含不正确的行

result_tasks += mult_tasks_line(ch, str_numbers),"\n"

result_tasks 是一个字符串，在这里它与一个元组连接（它们由带或不带括号的逗号构成）mult_tasks_line(ch, str_numbers), "\n" 导致异常。

您需要将此行修改为

result_tasks += mult_tasks_line(ch, str_numbers) + "\n"

代码中的另一个问题是 mult_tasks_line 没有 return 值，只是打印出来。你想做这样的事情

def mult_tasks_line(first_num, str_numbers):
    result = []
    for ch in str_numbers:
        result.append('({}*{})'.format(first_num, ch))
    return ' '.join(result)

封装和泛化函数 def mult_tasks(str_numbers) in python

Encapsulation and generalization function def mult_tasks(str_numbers) in python

python

iteration

encapsulation

generalization