迭代多个数组来执行任务?

时间:2015-07-01 09:19:43

标签: python arrays numpy iteration

我有9个数组,每个数组包含19个值。

让我们说它们是a1,a2,a3,a4,a5,a6,a7,a8,a9(每个a1,a2 ... a9各包含19个值),让我们称之为 a数组。

我还有9个数组,每个数组包含19个值。

我们说它们是b1,b2,b3,b4,b5,b6,b7,b8,b9(每个b1,b2 ... b9各包含19个值),让我们称它们为 b数组。

我现在想采用每个 a数组 第一个值和第一个值每个 b数组 ,将它们分开(a/b),这会给我一个新的数组,让我们来看看用{19}来说a/b。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a1b1),我可以使用zip,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)

如何在我冗长的情况下重复上述内容?

修改

我终于需要19个不同的标准差,即我计算第一个值,然后计算第二个值等等。

3 个答案:

答案 0 :(得分:6)

如果您将numpy用于std,为什么不使用>>> # 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) 的权力?

np.std

示例(包含>>> 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]) 的详细信息):

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答案 1 :(得分:0)

您可以将每个ab数组放在另一个数组中。所以它会是

a[0] = a1a[1] = a2等同等内容b

然后像:

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ab的长度相同,而a[0]b[0]等等也是相同的长度,则会有效。在你的情况下这是真的。

答案 2 :(得分:0)

不久前我写了一篇基本上可以满足你所要求的课程。唯一的技巧是你必须为每个数组传递一个迭代器列表。

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)

你可以通过

使用这段代码
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.