Python 从 for 循环中获取输出并将其分配到其他地方

Question

以可测试代码为例重新发布。

大家好。各位大侠能帮我解答一下吗？我一直在尝试获取 for 循环的输出并使用它来打印摘要页面。我希望 for 循环中的每次迭代都成为最后一次迭代旁边的一列。你能帮助实现这一目标吗？非常感谢您的帮助。

导入时间、re、集合、运算符

output_list = [['2016-07-12', 'Magazine', 'News Paper #2', 'Podcast', '1234567', '10-10-10-10', 'ABCDEFG', 'Zoo'],
['2016-07-12', 'Book', 'News Paper #2', 'Podcast', '1234567', '10-10-10-10', 'ABCDEFG', 'Zoo']]


def count_types():
    item_1 = mm_counts(1)
    item_2 = mm_counts(4)
    item_3 = mm_counts(3)

def mm_counts(a):
    r = []
    for i in output_list:
        x = (i[a])
        #x = (i[0] + ': ' + i[a])
        r.append(x)
    y = collections.Counter(r)
    #test_list = []
    for k, v in sorted(y.items(), key=operator.itemgetter(1), reverse=True):
        z = (str(k).ljust(5, ' ') + ' ' + (str(v).ljust(5, ' ')))
        print(z)   #<--- I want to print this column and iterate next columns next to each other.

count_types()

电流输出：

Magazine 1    
Book  1    
1234567 2    
Podcast 2   

期望的输出：

Magazine 1  1234567 2
Book     1  Podcast 2

Answer 1

选择您想要的列数，打印在一行上，每个元素后有一个制表符，每次为每行打印所需数量的元素时转到下一行。

如果您的目标是拥有两列，请将desiredColumns 设置为两列。如果您的目标是两行，请将 desiredColumns 设置为您将要打印的总数量除以 2，然后四舍五入。

import time, re, collections, operator

output_list = [
    ['2016-07-12', 'Magazine', 'News Paper #2', 'Podcast', '1234567', '10-10-10-10', 'ABCDEFG', 'Zoo'],
    ['2016-07-12', 'Book', 'News Paper #2', 'Podcast', '1234567', '10-10-10-10', 'ABCDEFG', 'Zoo']]


def count_types():
    item_1 = mm_counts(1)
    item_2 = mm_counts(4)
    item_2 = mm_counts(3)

def mm_counts(a):
    r = []
    for i in output_list:
        x = (i[a])
        #x = (i[0] + ': ' + i[a])
        r.append(x)
    y = collections.Counter(r)
    test_list = []

    wordsInColumn = 0
    desiredColumns = 2

    for k, v in sorted(y.items(), key=operator.itemgetter(1), reverse=True):
        z = (str(k).ljust(5, ' ') + ' ' + (str(v).ljust(5, ' ')))
        print (z, end = '\t')
        wordsInColumn += 1
        if wordsInColumn % desiredColumns == 0:
          print('')

count_types()

#    output:
#    Magazine 1     Book  1     
#    1234567 2      Podcast 2