如何在Python中遍历一列? [英] How to loop through a column in Python?

问题描述

我已经看到了有关此问题的答案，但没有人帮助我.有些人使用了 numpy ，有些人使用了其他有助于Python更简单的平台来回答问题.我不需要这些类型的东西，我希望使用简单的Python而不导入库或其他任何东西.

I have seen answers about this question but no one helped me. Some used numpy, and some people answered using other platforms that help Python to be simpler. I don't want these type of things, I want with the simple Python without importing libraries or anything more.

让我们说:我想做一个方法来检查2D数组中是否至少有一个列具有相同的值. 例如:

Let's say: I would want to do a method that checks if there's at least one column in the 2D array that the column has the same values. For example:

arr = [[2,0,3],[4,2,3],[1,0,3]]

将arr发送到我的方法将返回True，因为在第三列中每个术语中都有数字3.

Sending arr to my method would return True because in the third column there is in each term the number 3.

我将如何编写此方法?如何遍历2D数组中的每一列?

How would I write this method? How do I loop through each column in the 2D array?

推荐答案

遍历列

如何遍历2D数组中的每一列?

How do I loop through each column in the 2D array?

为了遍历每一列，只需遍历转置矩阵(转置矩阵只是一个新矩阵原来矩阵的行现在是列，反之亦然).

In order to loop through each column just loop through the transposed matrix (a transposed matrix is just a new matrix where the rows of original matrix are now columns and vice-versa).

# zip(*matrix) generates a transposed version of your matrix
for column in zip(*matrix): 
    do_something(column)

您提出的问题/示例的答案

我想做一个检查至少一列的方法 在二维数组中该列具有相同的值

I would want to do a method that checks if there's at least one column in the 2D array that the column has the same values

常规方法:

def check(matrix):
    for column in zip(*matrix):
        if column[1:] == column[:-1]:
            return True
    return False

单线:

arr = [[2,0,3],[4,2,3],[1,0,3]]
any([x[1:] == x[:-1] for x in zip(*arr)])

说明:

arr = [[2,0,3],[4,2,3],[1,0,3]]
# transpose the matrix
transposed = zip(*arr) # transposed = [(2, 4, 1), (0, 2, 0), (3, 3, 3)]
# x[1:] == x[:-1] is a trick.
# It checks if the subarrays {one of them by removing the first element (x[1:])
# and the other one by removing the last element (x[:-1])} are equals.
# They will be identical if all the elements are equal. 
equals = [x[1:] == x[:-1] for x in transposed] # equals = [False, False, True]
# verify if at least one element of 'equals' is True
any(equals) # True

更新01

@BenC写道:

Update 01

@BenC wrote:

您也可以跳过列表理解中的[]，以便任何 只是得到了可以提前停止一次/如果返回的发电机 假"

"You could also skip the [] around the list comprehension so that any just gets a generator that can be stopped early once/if it returns false"

如此:

arr = [[2,0,3],[4,2,3],[1,0,3]]
any(x[1:] == x[:-1] for x in zip(*arr))

更新02

您还可以使用集合(与@HelloV的答案合并).

Update 02

You could also use sets (merged with the answer of @HelloV).

单线:

arr = [[2,0,3],[4,2,3],[1,0,3]]
any(len(set(x))==1 for x in zip(*arr))

常规方法:

def check(matrix):
    for column in zip(*matrix):
        if len(set(column)) == 1:
            return True
    return False

集合没有重复的元素，因此，如果将列表转换为集合set(x)，则任何重复的元素都会消失，因此，如果所有元素相等，则结果集的长度等于一个len(set(x))==1.

A set does not have repeated elements, so if you transform a list into a set set(x) any duplicated element goes away, so, if all elements are equals, the lenght of resulting set is equal to one len(set(x))==1.

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