How to loop through a column in Python?

问题

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.

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]]

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

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

回答1:

Loop through column

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

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)

An answer to your proposed problem/example

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

General method:

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

One-liner:

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

Explanation:

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

Update 01

@BenC wrote:

"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"

so:

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

Update 02

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

One-liner:

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

General method:

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

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.

回答2:

A simple example, without adding the complexity of list comprehensions and the zip function is the following:

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

def check_column_equals_index(colum):
    for row in arr:
        if row[colum-1] != colum:
            return False
    return True

print check_column_equals_index(3)

which will output True if the 3th column is equal to 3 for every row.

Nonetheless, as you might have read in some other related discussions, it might be worth considering to use Numpy or Pandas.

回答3:

1 in [len(set(i)) for i in zip(*arr)]

回答4:

Loop through a column in 2D list and no imports? How about extract columns elements to auxiliary list and then compare among themselves. You can control the process in this synoptical function:

def checkcolumn(colnum, arrex=[]):
    for i in range(len(arr)):
        arrex.append(arr[i][colnum])

    if arrex.count(arrex[0]) == len(arrex):
        return True
    else:
        return False

print checkcolumn(1) 

Seems easiest and pretty didactic to me. Other ways of comparing elements in column can be found at: https://www.csestack.org/python-check-if-all-elements-in-list-are-same/

来源：https://stackoverflow.com/questions/34386476/how-to-loop-through-a-column-in-python