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Python &引用;至于;循环x值?_Python_Loops - Fatal编程技术网

Python &引用;至于;循环x值?

Python &引用;至于;循环x值?,python,loops,Python,Loops,我试图让我的第二个数组以增量方式打印x,但一旦大小写值发生变化,就会重置为零 代码: 电流输出: 2017001677 0 2017001677 1. 2017001621 2. 2017001621 3. 期望输出: 2017001677 0 2017001677 1. 2017001621 0 2017001621 1. 如何完成此操作?您可以通过跟踪上次看到的值(从无开始)来重置数组中唯一值的计数器 array = ['2017001677', '2017001677', '20170

我试图让我的第二个数组以增量方式打印
x
,但一旦大小写值发生变化,就会重置为零

代码:

电流输出:

2017001677
0
2017001677
1.
2017001621
2.
2017001621
3.
期望输出:

2017001677
0
2017001677
1.
2017001621
0
2017001621
1.

如何完成此操作?

您可以通过跟踪上次看到的值(从
无开始)来重置
数组中唯一值的计数器

array = ['2017001677', '2017001677', '2017001621', '2017001621']
array2 = ['2017001677', '2017001621']

last_case = None
for case in array:
    if case != last_case:
        x = 0
        last_case = case

    for case2 in array2:
        if(case == case2):
            print(case2)
            print(x)
            x = x + 1

如果您的输入是
['2017001677','2017001621','2017001677','2017001621']
,您的输出是什么<代码>0 0 0 0
0 0 1 1
?这将是元素的正确x值,但我需要数组处于原始顺序数组不打印东西…因此我不清楚你在问什么。这很接近,但结果是错误的,我希望重复的数组是增量的。你的结果:2017001677 0 2017001677 0 2017001621 1 2017001621 1我想要什么:2017001677 0 2017001677 1 2017001621 0 2017001621 1 1
array = ['2017001677', '2017001677', '2017001621', '2017001621']
array2 = ['2017001677', '2017001621']

last_case = None
for case in array:
    if case != last_case:
        x = 0
        last_case = case

    for case2 in array2:
        if(case == case2):
            print(case2)
            print(x)
            x = x + 1