如何使用Python处理带有中断的双数组? [英] How to process double array with break using Python?

问题描述

我想处理一个数组.这是我以前的问题.如何在Python中处理中断数组?

I would like to process an array. Here is my previous question. How to process break an array in Python?

但是我还有另一种情况，我参考了这个答案，但是我仍然可以得到我的期望结果.就是这种情况

But I have another case and I refer to that answer, but I still can get my expectation result. Here is the case

Folder = "D:\folder"
Name = ['gadfg5', '546sfdgh']
Ver = [None, 'hhdt5463']
Default = {'gadfg5': '6754435'}

情况二

Folder = "D:\folder"
Name = ['gadfg5', '546sfdgh']
Ver = [None, 'None']
Default = {'gadfg5': '6754435', '546sfdgh': '98769786'}

情况3

Folder = "D:\folder"
Name = ['gadfg5', '546sfdgh']
Ver = [g675436g, 'hhdt5463']
Default = {}

这是我尝试过的:

for dn, dr, key in zip(Name, Ver, Default):
        if dr is None:
            Path = os.path.join(Folder, dn, Default[key])
            print("None",Path)
        else:
            Path = os.path.join(Folder, dn, dr)
            print("Not None", Path)

CASE 3 的输出为 empty .但我期望输出应该是:

The output of CASE 3 is empty. But my expectation the output supposed to be:

D:\folder\gadfg5\g675436g
D:\folder\546sfdgh\hhdt5463

CASE 2 的输出符合我的期望，是正确的:

The output of CASE 2 is correct as my expectation which is:

D:\folder\gfg\6754435
D:\folder\546sfdgh\98769786

案例1 的输出仅返回一条路径，如下所示:

The output of CASE 1 only returns one path, like this:

D:\folder\gadfg5\6754435

但是我期望输出应该是这样的:

But my expectation the output supposed to be like this:

 D:\folder\gadfg5\6754435
 D:\folder\546sfdgh\hhdt5463

任何人都可以给我一个主意.谢谢

Anyone can give me an idea, please. Thanks

推荐答案

如果一个是 Name 和 Ver 包含两个元素而字典 Default 只有一个，因此当您压缩它们时，输出将只有一个元组:('gadfg5'，None，'gadfg5')

what happens in case one is Name and Ver has two elements while dictionary Default has just one so when you zip them the output will have just one tuple: ('gadfg5', None, 'gadfg5')

由于 Default 是一本字典，并且其键是 Names 的元素，所以我们不必压缩它们中的三个，而是尝试:

Since the Default is a dictionary and its keys are the elements of Names we don't have to zip the three of them, instead try:

for dn, dr in zip(Name, Ver):
    if dr is None:
        Path = os.path.join(Folder, dn, Default[dn])
        print("None", Path)
    else:
        Path = os.path.join(Folder, dn, dr)
        print("Not None", Path)

相同的逻辑适用于情况3

The same logic applies to case 3

但是这里有一个问题，让我用情况4进行演示:

But there is an issue here let me demonstrate with another scenario, case 4:

输入:

Folder = "D:\Folder"
Name = ['gadfg5', '546sfdgh']
Ver = [None, 'hhdt5463']
Default = {}

此处dict < Default 为空， Ver 具有None元素.这将在 Default [dn] 处引发Key错误.因此，我们也如下进行检查:

Here dict Default is empty and Ver has a None element. This will throw a Key error at Default[dn]. So lets put a check for that too as follows:

for dn, dr in zip(Name, Ver):
    if dr is None:
        if dn in Default:  # check if Default contains the key dn
            Path = os.path.join(Folder, dn, Default[dn])
            print("None", Path)
        else:
            print('no default path for ', dn)
    else:
        Path = os.path.join(Folder, dn, dr)
        print("Not None", Path)

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