CODEWITHSUNDEEP
python
Kristy
Kristy

Reputation: 25

python using arrays to make a matrix

I am trying to make a matrix out of arrays

Suppose My inputs are two arrays 

       my_column = [['a','2'] ['k','34','2'] ['d','e','5']]
       my_row = ['a' '2' 'k' '34' 'd' 'e' '5']

I want an output like this

             #a    #2     #k     #34    #d    #e    #5
      #a,2    1     1     0       0      0     0     0
     #k,34,2  0     1     1       1      0     0     0
     #d,e,5   0     0     0       0      1     1     1

However, I am representing this output in a 2D list. My desired output is this

      output_matrix = [['1', '1', '0', '0', '0', '0', '0'],['0', '1', '1', '1', '0', '0', '0'],['0', '0', '0', '0', '1', '1', '1']]

I am matching my row and column and getting a matrix of 1s if matched and 0 otherwise. #stuff is just a comment for better understanding.

       output_matrix = [[]]
       for i in output_matrix:
          for j in i:
              if my_row[i] == my_column[i][j]:
                 output_matrix.append(1)
              else:
                  output_matrix.append(0)

I am trying to create a new 2D list and store values in that.However, my entire approach seems to be wrong as I get the output as just [ [ ] ]

Upvotes: 0

Views: 943

Answers (4)

user5592768
user5592768

Reputation:

Your default output is definitely correct. If you want specific output, you can code your own like:

# Cosmetic variables
padding = 8
truncating = 6

# Data prepared for printing.
my_column = [truncating*'-'] + ['a', '2', 'k', '34', 'd', 'e', '5']
my_row = [['a', '2'],[ 'k', '34', '2'],['d', 'e', '5']]
output_matrix = [['1', '1', '0', '0', '0', '0', '0'],['0', '1', '1', '1', '0', '0', '0'],['0', '0', '0', '0', '1', '1', '1']]


for element in my_column:
    print('#{:{}.{}}'.format(''.join(element), padding, truncating), end="")
else:
    print()

for index, list in enumerate(output_matrix):
    print('#{:{}.{}}'.format(','.join(my_row[index]), padding, truncating), end="")

    for element in list:
        print(' {:{}.{}}'.format(''.join(element), padding, truncating), end="")
    else:
        print()

Python 3.x

The result is close to your wish:

#------  #a       #2       #k       #34      #d       #e       #5 
#a,2      1        1        0        0        0        0        0
#k,34,2   0        1        1        1        0        0        0
#d,e,5    0        0        0        0        1        1        1

How can you do this by-self?

Website: Using % & .format()

Documentation: Format String Syntax

Cheer!

Upvotes: 0

DisappointedByUnaccountableMod
DisappointedByUnaccountableMod

Reputation: 6826

Given the input you provide (i had to add commas between the items in the lists), and the output you want, this does the trick:

my_column = [['a','2'], ['k','34','2'], ['d','e','5']]
my_row = ['a', '2', 'k', '34', 'd', 'e', '5']

output_matrix = []
for r in my_column:
    print "r=",r
    output_matrix.append([])
    for c in my_row:
        print "c=",c
        if c in r:
            output_matrix[-1].append(1)
        else:
            output_matrix[-1].append(0)
print output_matrix

Output is:

[[1, 1, 0, 0, 0, 0, 0], [0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1]]

BTW I would say your names my_column and my_row are the wrong way round - i.e. the column should be the inner lists, the row should be the outer lists.

Another way of doing this would be to first create an array (a list of lists) of the requisite size with all values 0, then go in and write the 1s where there are matches.

# construct a matrix with as many columns as my_row and as many rows as my_column
output_matrix1 = [[0 for c in my_row] for r in my_column]
for x,colvalue in enumerate( my_row ):
    for y, rowvalue in enumerate( my_column ):
        if colvalue in rowvalue:
            output_matrix1[y][x] = 1
print output_matrix1

Upvotes: 0

Ismail Moghul
Ismail Moghul

Reputation: 2984

This uses numpy...

import numpy as np
my_column = [['a','2'], ['k','34','2'], ['d','e','5']]
my_row = ['a', '2', 'k', '34', 'd', 'e', '5']

results = np.zeros((len(my_column), len(my_row)))

for i in range(0, len(my_column) ):
    for j in my_column[i]:
        for ind, x in enumerate(my_row):
            if x == j: 
                results[i,ind] = 1

results

array([[ 1.,  1.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  1.,  1.,  1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  1.,  1.,  1.]])

Upvotes: 1

PanGalactic
PanGalactic

Reputation: 72

my_column = [['a','2'],['k','34','2'],['d','e','5']]
my_row = ['a', '2', 'k', '34', 'd', 'e', '5']

output = []
for elem in my_column:
    output_row = []
    for val in my_row:
        if val in elem:
            output_row.append(1)
        else:
            output_row.append(0)
    output.append(output_row)
print output

This will print out:

[[1, 1, 0, 0, 0, 0, 0], [0, 1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1]]

Upvotes: 0

Related Questions