迭代多个数组来执行任务？

Question

我有 9 个数组，每个数组包含 19 个值。

假设它们是a1,a2,a3,a4,a5,a6,a7,a8,a9（每个 a1、a2...a9 包含 19 个值），我们称它们为 a 数组。

我还有 9 个数组，每个数组包含 19 个值。

假设它们是b1,b2,b3,b4,b5,b6,b7,b8,b9（每个 b1、b2...b9 包含 19 个值），我们称它们为 b 数组。

我现在想获取每个 a数组的第一个值和每个第一个值的 b 数组，将它们划分为(a/b)，这将给我一个新数组，假设a/b 有19 个值。然后我使用numpy.std 计算这 19 个值的标准差。

然后我想再次遍历这些数组，但是这次是每个数组的第二个值，依此类推，直到最后一个（第 19 个）值并执行上述操作。

如果我只有 2 个数组（比如 a1 和 b1），我可以使用 zip，例如：

div_array = [] # The empty array that will have the divided values
for a,b in zip(a1,b1):
    div = a/b
    div_array.append(div)

std = np.std(div_array)

如何在我的冗长案例中重复上述内容？

编辑：

我最终需要 19 个不同的标准差， 即我为第一个值计算它，然后是第二个值，依此类推..

Answer 1

如果你用std除法，为什么不使用numpy的力量呢？

>>> # You can create these array in a loop if you want
>>> a = np.array([a1, a2, a3, ..., a9])
>>> b = np.array([b1, b2, b3, ..., b9]) 
>>> c = np.std(a / b, 0)

示例（包含有关np.std 的详细信息）：

>>> a1 = np.array([1, 2, 3])
>>> a2 = np.array([2, 3, 4])
>>> a  = np.array([a1, a2])
>>> a
array([[1, 2, 3],
       [2, 3, 4]])
>>> b1 = np.array([10, 100, 1000])
>>> b2 = np.array([20, 200, 2000])
>>> b  = np.array([b1, b2])
>>> b
array([[10, 100, 1000], 
       [20, 200, 2000]])
>>> a/b
array([[0.1,  0.02, 0.003],
       [0.1, 0.015, 0.002]])
>>> np.std(a/b)             # The standard deviation of the whole matrix
0.04289... 
>>> np.std(a/b, 0)          # The standard deviation of each column
array([0, 0.0025, 0.0005])
>>> np.std(a/b, 1)          # The standard deviation of each row
array([0.04229263, 0.04345879])

Answer 2

所以不久前，我写了一个基本上可以满足您要求的课程。唯一的技巧是你必须向类传递每个数组的迭代器列表。

column_iter_traversal.py

"""
This code will take in a iterator of iterators. You can view the first
iterator as the rows of a graph (a matrix being a specific case of graphs)
and each iterable giving you the columns (or nodes) of that graph.

so if you have a graph
[[1, 2],
[3],
[4, 5]]

we'd expect the iterator to return [1, 3, 4, 2, 5]
"""

class ColumnTraversalIter():
    """
    This is a class which is used to contain the currently travered state.
    This is the class which defines the returned object for
    column_traversal_iter. The iter that function returns is an instance of
    this class.
    """
    def __init__(self, iter_of_iters):
        # Build a list of iterators
        self.iter_list = []
        for it in iter_of_iters:
            self.iter_list.append(it)
        self.current_iter_index = 0

    def __iter__(self):
        return self

    def __next__(self):
        # Get the next value from the current iterator
        try:
            return_val = next(self.iter_list[self.current_iter_index])
            self.current_iter_index = self._increment_index(
                self.current_iter_index,
                len(self.iter_list))
            return return_val
        except StopIteration:
            # When we run into a stop iteration we know that the current
            # iterator is out of values. Remove the current iterator from
            # the iterator list.
            del self.iter_list[self.current_iter_index]
            # If we are out of iterators it's time to raise StopIteration
            if len(self.iter_list) == 0:
                raise StopIteration
            else:
                # Otherwise, set the current_iter_index and recall next
                self.current_iter_index = self._increment_index(
                    self.current_iter_index,
                    len(self.iter_list))
                return self.__next__()
        except IndexError:
            # Someone called __next__ when there aren't any iterators left in
            # the iter_list.
            raise StopIteration

    @staticmethod
    def _increment_index(iter_index, wrap_length):
        if iter_index + 1 > wrap_length:
            print("returning 0")
            return 0
        else:
            print("returning {}".format(iter_index + 1))
            return iter_index + 1

def column_traversal_iter(iter_of_iters):
    """

    args:
        iterator: a iterator of iterators. If there aren't any iterators or
                  there are non iterator elements this will explode.
    returns a COlumnTraversalIter
    """
    return ColumnTraversalIter(iter_of_iters)

tests.py

import unittest
from column_traversal import column_traversal_iter

class TestBruteforceImplemetation(unittest.TestCase):

    def test_no_iters(self):
        test_iter = iter([])
        column_iter = column_traversal_iter(test_iter)
        with self.assertRaises(StopIteration):
            next(column_iter)

    def test_iter_of_one_empty_iter(self):
        """
        One empty iter and many empty iters should hit one stop iteration.
        """
        test_iter = iter([iter([])])
        column_iter = column_traversal_iter(test_iter)
        with self.assertRaises(StopIteration):
            next(column_iter)

    def test_iter_of_many_empty_iter(self):
        """
        One empty iter and many empty iters should hit one stop iteration.
        """
        test_iter = iter([iter([]), iter([]), iter([])])
        column_iter = column_traversal_iter(test_iter)
        with self.assertRaises(StopIteration):
            next(column_iter)

    def test_iter_simple_one_by_one_matrix(self):
        """
        One empty iter and many empty iters should hit one stop iteration.
        """
        test_iter = iter([iter([1]), iter([2]), iter([3])])
        column_iter = column_traversal_iter(test_iter)
        expected_traversal = [1, 2, 3]
        for actual_value, expected_value in zip(column_iter, expected_traversal):
            self.assertEqual(actual_value, expected_value)

        # Check to make sure there's a stop iteration.
        with self.assertRaises(StopIteration):
            next(column_iter)

    def test_iter_simple_jagged_graph(self):
        """
        One empty iter and many empty iters should hit one stop iteration.
        """
        test_iter = iter([iter([1]), iter([2, 4]), iter([3])])
        column_iter = column_traversal_iter(test_iter)
        expected_traversal = [1, 2, 3, 4]
        for actual_value, expected_value in zip(column_iter, expected_traversal):
            self.assertEqual(actual_value, expected_value)

        # Check to make sure there's a stop iteration.
        with self.assertRaises(StopIteration):
            next(column_iter)

    def test_iter_simple_two_by_two_matrix(self):
        """
        One empty iter and many empty iters should hit one stop iteration.
        """
        test_iter = iter([iter([1, 4]), iter([2, 5]), iter([3, 6])])
        column_iter = column_traversal_iter(test_iter)
        expected_traversal = [1, 2, 3, 4, 5, 6]
        for actual_value, expected_value in zip(column_iter, expected_traversal):
            self.assertEqual(actual_value, expected_value)

        # Check to make sure there's a stop iteration.
        with self.assertRaises(StopIteration):
            next(column_iter)

    def test_iter_one_iter_is_blank(self):
        """
        One empty iter and many empty iters should hit one stop iteration.
        """
        test_iter = iter([iter([1, 3]), iter([2, 4]), iter([])])
        column_iter = column_traversal_iter(test_iter)
        expected_traversal = [1, 2, 3, 4]
        for actual_value, expected_value in zip(column_iter, expected_traversal):
            self.assertEqual(actual_value, expected_value)

        # Check to make sure there's a stop iteration.
        with self.assertRaises(StopIteration):
            next(column_iter)

你会使用这个代码

divisor_iter_list = []
for a_list in a_lists:
    divisor_iter_list.append(iter(a_list))

dividend_iter_list = []
for b_list in b_lists:
    divident_iter_list.append(iter(b_list))


divisor_iter = ColumnTraversalIter(divisor_iter_list)
dividend_iter = ColumnTraversalIter(divident_iter_list)
for divisor, dividend in zip(divisor_iter, dividend_iter):
    # Do calculations.

Answer 3

您可以将每个 a 和 b 数组放在另一个数组中。所以会是

a[0] = a1、a[1] = a2 等b 也是如此

然后是这样的：

for i in range(length(a)):
    div_array = []
    for j in range(length(a[i])):
        div = a[i][j]/b[i][j]
        div_array.append(div)

    std.append(np.std(div_array))

然后你有一个数组std，其中包含你想要的所有值。

当然，如果a 和b 的长度相同，并且a[0] 和b[0] 等的长度也相同，这将起作用。你的情况是这样。