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      如何迭代这个 n 维数据集?

      时间:2023-10-19
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                本文介绍了如何迭代这个 n 维数据集?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                问题描述

                我有一个 dataset,它有 4 个维度(目前...),我需要对其进行迭代.

                I have a dataset which has 4 dimensions (for now...) and I need to iterate over it.

                要访问 dataset 中的值,我这样做:

                To access a value in the dataset, I do this:

                value = dataset[i,j,k,l]
                

                现在,我可以获得 datasetshape:

                Now, I can get the shape for the dataset:

                shape = [4,5,2,6]
                

                shape中的值代表维度的长度.

                The values in shape represent the length of the dimension.

                在给定维数的情况下,我如何迭代数据集中的所有元素?这是一个例子:

                How, given the number of dimensions, can I iterate over all the elements in my dataset? Here is an example:

                for i in range(shape[0]):
                    for j in range(shape[1]):
                        for k in range(shape[2]):
                            for l in range(shape[3]):
                                print('BOOM')
                                value = dataset[i,j,k,l]
                

                将来,shape 可能会发生变化.例如,shape 可能有 10 个元素,而不是当前的 4 个.

                In the future, the shape may change. So for example, shape may have 10 elements rather than the current 4.

                在 Python 3 中是否有一种简洁明了的方式来做到这一点?

                Is there a nice and clean way to do this with Python 3?

                推荐答案

                你可以使用 itertools.product 迭代 笛卡尔积1 个值(在本例中为索引):

                You could use itertools.product to iterate over the cartesian product 1 of some values (in this case the indices):

                import itertools
                shape = [4,5,2,6]
                for idx in itertools.product(*[range(s) for s in shape]):
                    value = dataset[idx]
                    print(idx, value)
                    # i would be "idx[0]", j "idx[1]" and so on...
                

                <小时>

                但是,如果它是您想要迭代的 numpy 数组,可能更容易使用 np.ndenumerate:


                However if it's a numpy array you want to iterate over, it could be easier to use np.ndenumerate:

                import numpy as np
                
                arr = np.random.random([4,5,2,6])
                for idx, value in np.ndenumerate(arr):
                    print(idx, value)
                    # i would be "idx[0]", j "idx[1]" and so on...
                

                <小时>

                1 您要求澄清 itertools.product(*[range(s) for s in shape]) 的实际作用.所以我会更详细地解释它.


                1 You asked for clarification what itertools.product(*[range(s) for s in shape]) actually does. So I'll explain it in more details.

                例如你有这个循环:

                for i in range(10):
                    for j in range(8):
                        # do whatever
                

                这也可以用 product 写成:

                for i, j in itertools.product(range(10), range(8)):
                #                                        ^^^^^^^^---- the inner for loop
                #                             ^^^^^^^^^-------------- the outer for loop
                    # do whatever
                

                这意味着 product 只是减少 independant for 循环数量的便捷方式.

                That means product is just a handy way of reducing the number of independant for-loops.

                如果您想将可变数量的 for-loops 转换为 product,您基本上需要两个步骤:

                If you want to convert a variable number of for-loops to a product you essentially need two steps:

                # Create the "values" each for-loop iterates over
                loopover = [range(s) for s in shape]
                
                # Unpack the list using "*" operator because "product" needs them as 
                # different positional arguments:
                prod = itertools.product(*loopover)
                
                for idx in prod:
                     i_0, i_1, ..., i_n = idx   # index is a tuple that can be unpacked if you know the number of values.
                                                # The "..." has to be replaced with the variables in real code!
                     # do whatever
                

                相当于:

                for i_1 in range(shape[0]):
                    for i_2 in range(shape[1]):
                        ... # more loops
                            for i_n in range(shape[n]):  # n is the length of the "shape" object
                                # do whatever
                

                这篇关于如何迭代这个 n 维数据集?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!

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