如何堆叠索引和给定值 [英] How to stack indices and given values

问题描述

seq1_values = [5, 40, 180, 13, 30, 20, 25, 24, 29,  31,  54, 46,  42, 50,  67, 17,
           76, 33, 84, 35, 100, 37, 110, 32, 112, 15, 123, 3, 130, 42]

def get_num_values():
    global seq1_values
    return len( seq1_values )

def get_values():
    global seq1_values
    for value in seq1_values:
        yield value
    yield None

def next_value():
    return next( next_value.values )
next_value.values = get_values()

def main():
    numbers = []
    value = next_value()
    while value is not None:
        numbers.append(value)
        input_size = len(numbers)-1
        for i in range(input_size):
            print("|", str(input_size).rjust(4), end="")
        for j in range(input_size):
            print("|", str(value).rjust(4), end="")
            value = next_value()
main()

输出应该看起来像这样:

Output is supposed to look like this:

|    0|    1|    2|    4|    5|    6|    7|    8|    9|    10|    11|    12|    13|    14|    15|    16|    17|    18|    19|    20|    21|    22|    23|    24|    25|   26|    27|    28|    29|    30|

|    5|    40|    180|    13|    30|    20|    25|    24|    29|    31|    54|    46|    42|    50|    67|    17|    76|    33|    84|    35|    100|    37|    110|    32|    112|   15|    123|    3|    130|    42|

推荐答案

您正在使它变得比所需的复杂得多.它闻起来像是作业，所以我将帮助您进行堆栈部分的工作，但是在此之后仍然需要证明它是正确的，并且这不会在末尾添加竖线字符，因此您必须在自己的代码上弄清楚这一点.自己的:

You are making this a lot more complex than it needs to be. It smells like homework, so I am going to help you along for the stacking portion, but it still needs to be justified after this, and this does not add a pipe character at the end, so you will have to figure that out on your own:

在Python3中，您需要的只是这样的东西:像您需要的那样将其堆叠:

Something like this is all that you need for this in Python3 for stacking it like you need it:

for idx in range(0, len(seq1_values)):
    if idx == len(values) - 1:
        print('| ', idx)
    else:
        print('| ', idx, end='')

for val in len(seq1_values):
    print('| ', val, end='')

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