Python - Simple Question: Numpy Matrix, Loop

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0
down vote

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I am fairly new to python and I am trying to get around understanding the following code:

import numpy as np

n=4

matrix=np.zeros((n,n))

for j in range (0,n):

    for i in range (n-1,n-j-2,-1):

        matrix[i,j]=2*n-i-j-1

print (matrix)

I would greatly appreciate if someone could please help me understand how each line executes and how the code is revaluated with the loop.

Thanks in advance!

edited Nov 22 at 0:41

asked Nov 21 at 23:28

Aylín Pérez

In short: It creates a 4 by 4 array where each cell in the lower left triangle has the value i - j + 1 where i is the row index and j is the column index. You need to be more specific about what you don't understand if that does not solve the issue.
– timgeb
Nov 21 at 23:30

Thanks for your prompt reply! My question is how do I know that the lower left triangle is the one where the matrix formula executes. Sorry if it is way too obvious, but I am a complete begginner
– Aylín Pérez
Nov 21 at 23:33

The outer for loop iterates over the row indices. The inner for loop iterates over the columns, but only up to the column number that is equal to the current row number.
– timgeb
Nov 21 at 23:34

1

Thank you very much!!
– Aylín Pérez
Nov 21 at 23:47

add a comment |

up vote
0
down vote

favorite

I am fairly new to python and I am trying to get around understanding the following code:

import numpy as np

n=4

matrix=np.zeros((n,n))

for j in range (0,n):

    for i in range (n-1,n-j-2,-1):

        matrix[i,j]=2*n-i-j-1

print (matrix)

I would greatly appreciate if someone could please help me understand how each line executes and how the code is revaluated with the loop.

Thanks in advance!

edited Nov 22 at 0:41

asked Nov 21 at 23:28

Aylín Pérez

In short: It creates a 4 by 4 array where each cell in the lower left triangle has the value i - j + 1 where i is the row index and j is the column index. You need to be more specific about what you don't understand if that does not solve the issue.
– timgeb
Nov 21 at 23:30

Thanks for your prompt reply! My question is how do I know that the lower left triangle is the one where the matrix formula executes. Sorry if it is way too obvious, but I am a complete begginner
– Aylín Pérez
Nov 21 at 23:33

The outer for loop iterates over the row indices. The inner for loop iterates over the columns, but only up to the column number that is equal to the current row number.
– timgeb
Nov 21 at 23:34

1

Thank you very much!!
– Aylín Pérez
Nov 21 at 23:47

add a comment |

up vote
0
down vote

favorite

I am fairly new to python and I am trying to get around understanding the following code:

import numpy as np

n=4

matrix=np.zeros((n,n))

for j in range (0,n):

    for i in range (n-1,n-j-2,-1):

        matrix[i,j]=2*n-i-j-1

print (matrix)

I would greatly appreciate if someone could please help me understand how each line executes and how the code is revaluated with the loop.

Thanks in advance!

edited Nov 22 at 0:41

asked Nov 21 at 23:28

Aylín Pérez

I am fairly new to python and I am trying to get around understanding the following code:

import numpy as np

n=4

matrix=np.zeros((n,n))

for j in range (0,n):

    for i in range (n-1,n-j-2,-1):

        matrix[i,j]=2*n-i-j-1

print (matrix)

I would greatly appreciate if someone could please help me understand how each line executes and how the code is revaluated with the loop.

Thanks in advance!

python python-2.7 loops numpy matrix

edited Nov 22 at 0:41

asked Nov 21 at 23:28

Aylín Pérez

edited Nov 22 at 0:41

asked Nov 21 at 23:28

Aylín Pérez

edited Nov 22 at 0:41

asked Nov 21 at 23:28

Aylín Pérez

asked Nov 21 at 23:28

Aylín Pérez

asked Nov 21 at 23:28

Aylín Pérez

In short: It creates a 4 by 4 array where each cell in the lower left triangle has the value i - j + 1 where i is the row index and j is the column index. You need to be more specific about what you don't understand if that does not solve the issue.
– timgeb
Nov 21 at 23:30

Thanks for your prompt reply! My question is how do I know that the lower left triangle is the one where the matrix formula executes. Sorry if it is way too obvious, but I am a complete begginner
– Aylín Pérez
Nov 21 at 23:33

The outer for loop iterates over the row indices. The inner for loop iterates over the columns, but only up to the column number that is equal to the current row number.
– timgeb
Nov 21 at 23:34

1

Thank you very much!!
– Aylín Pérez
Nov 21 at 23:47

add a comment |

In short: It creates a 4 by 4 array where each cell in the lower left triangle has the value i - j + 1 where i is the row index and j is the column index. You need to be more specific about what you don't understand if that does not solve the issue.
– timgeb
Nov 21 at 23:30

Thanks for your prompt reply! My question is how do I know that the lower left triangle is the one where the matrix formula executes. Sorry if it is way too obvious, but I am a complete begginner
– Aylín Pérez
Nov 21 at 23:33

The outer for loop iterates over the row indices. The inner for loop iterates over the columns, but only up to the column number that is equal to the current row number.
– timgeb
Nov 21 at 23:34

1

Thank you very much!!
– Aylín Pérez
Nov 21 at 23:47

In short: It creates a 4 by 4 array where each cell in the lower left triangle has the value i - j + 1 where i is the row index and j is the column index. You need to be more specific about what you don't understand if that does not solve the issue.
– timgeb
Nov 21 at 23:30

Thanks for your prompt reply! My question is how do I know that the lower left triangle is the one where the matrix formula executes. Sorry if it is way too obvious, but I am a complete begginner
– Aylín Pérez
Nov 21 at 23:33

The outer for loop iterates over the row indices. The inner for loop iterates over the columns, but only up to the column number that is equal to the current row number.
– timgeb
Nov 21 at 23:34

Thank you very much!!
– Aylín Pérez
Nov 21 at 23:47

add a comment |

2 Answers
2

active

oldest

votes

up vote
2
down vote

You can add the following print statement, and the loop will explain itself at each iteration:

n=4

matrix=np.zeros((n,n))

for i in range (0,n):

    for j in range(0,i+1):

        print(f'inserting {i-j+1} into the matrix at row index {i}, columns index {j}')

        matrix[i,j]=i-j+1

When you run it, you get this output:

inserting 1 into the matrix at row index 0, columns index 0

inserting 2 into the matrix at row index 1, columns index 0

inserting 1 into the matrix at row index 1, columns index 1

...

inserting 3 into the matrix at row index 3, columns index 1

inserting 2 into the matrix at row index 3, columns index 2

inserting 1 into the matrix at row index 3, columns index 3

And your matrix is populated as before:

>>> matrix

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

       [2., 1., 0., 0.],

       [3., 2., 1., 0.],

       [4., 3., 2., 1.]])

Just for reference:

>>> matrix

array([[1., 0., 0., 0.],   #<- "row" index 0

       [2., 1., 0., 0.],   #<- "row" index 1

       [3., 2., 1., 0.],   #<- "row" index 2

       [4., 3., 2., 1.]])  #<- "row" index 3



      # ^      ...  ^

      # "col" 0     "col" 3

answered Nov 21 at 23:33

sacul

29.7k41640

2

Thank you very much, this was really helpful!!
– Aylín Pérez
Nov 21 at 23:47

add a comment |

up vote
1
down vote

import numpy as np

n=4

We start by setting a 4x4 matrix with all coordinates set to null:

matrix=np.zeros((n,n))

We set new coordinate values by looping through rows and columns. First we loop through rows, from index 0 to n-1:

for i in range (0,n):

We next loop through columns. Now, notice that we only loop through those columns whose index is smaller than or equal to that of the current row (i.e., from 0 to i). This way we make sure that the values we set are on or below the diagonal of the matrix:

    for j in range(0,i+1):

Finally, we set the desired value for the current coordinate:

        matrix[i,j]=i-j+1

print(matrix)

answered Nov 21 at 23:46

DavidPM

18117

Thank you, really helpful!
– Aylín Pérez
Nov 21 at 23:53

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
2
down vote

You can add the following print statement, and the loop will explain itself at each iteration:

n=4

matrix=np.zeros((n,n))

for i in range (0,n):

    for j in range(0,i+1):

        print(f'inserting {i-j+1} into the matrix at row index {i}, columns index {j}')

        matrix[i,j]=i-j+1

When you run it, you get this output:

inserting 1 into the matrix at row index 0, columns index 0

inserting 2 into the matrix at row index 1, columns index 0

inserting 1 into the matrix at row index 1, columns index 1

...

inserting 3 into the matrix at row index 3, columns index 1

inserting 2 into the matrix at row index 3, columns index 2

inserting 1 into the matrix at row index 3, columns index 3

And your matrix is populated as before:

>>> matrix

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

       [2., 1., 0., 0.],

       [3., 2., 1., 0.],

       [4., 3., 2., 1.]])

Just for reference:

>>> matrix

array([[1., 0., 0., 0.],   #<- "row" index 0

       [2., 1., 0., 0.],   #<- "row" index 1

       [3., 2., 1., 0.],   #<- "row" index 2

       [4., 3., 2., 1.]])  #<- "row" index 3



      # ^      ...  ^

      # "col" 0     "col" 3

answered Nov 21 at 23:33

sacul

29.7k41640

2

Thank you very much, this was really helpful!!
– Aylín Pérez
Nov 21 at 23:47

add a comment |

up vote
2
down vote

You can add the following print statement, and the loop will explain itself at each iteration:

n=4

matrix=np.zeros((n,n))

for i in range (0,n):

    for j in range(0,i+1):

        print(f'inserting {i-j+1} into the matrix at row index {i}, columns index {j}')

        matrix[i,j]=i-j+1

When you run it, you get this output:

inserting 1 into the matrix at row index 0, columns index 0

inserting 2 into the matrix at row index 1, columns index 0

inserting 1 into the matrix at row index 1, columns index 1

...

inserting 3 into the matrix at row index 3, columns index 1

inserting 2 into the matrix at row index 3, columns index 2

inserting 1 into the matrix at row index 3, columns index 3

And your matrix is populated as before:

>>> matrix

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

       [2., 1., 0., 0.],

       [3., 2., 1., 0.],

       [4., 3., 2., 1.]])

Just for reference:

>>> matrix

array([[1., 0., 0., 0.],   #<- "row" index 0

       [2., 1., 0., 0.],   #<- "row" index 1

       [3., 2., 1., 0.],   #<- "row" index 2

       [4., 3., 2., 1.]])  #<- "row" index 3



      # ^      ...  ^

      # "col" 0     "col" 3

answered Nov 21 at 23:33

sacul

29.7k41640

2

Thank you very much, this was really helpful!!
– Aylín Pérez
Nov 21 at 23:47

add a comment |

up vote
2
down vote

You can add the following print statement, and the loop will explain itself at each iteration:

n=4

matrix=np.zeros((n,n))

for i in range (0,n):

    for j in range(0,i+1):

        print(f'inserting {i-j+1} into the matrix at row index {i}, columns index {j}')

        matrix[i,j]=i-j+1

When you run it, you get this output:

inserting 1 into the matrix at row index 0, columns index 0

inserting 2 into the matrix at row index 1, columns index 0

inserting 1 into the matrix at row index 1, columns index 1

...

inserting 3 into the matrix at row index 3, columns index 1

inserting 2 into the matrix at row index 3, columns index 2

inserting 1 into the matrix at row index 3, columns index 3

And your matrix is populated as before:

>>> matrix

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

       [2., 1., 0., 0.],

       [3., 2., 1., 0.],

       [4., 3., 2., 1.]])

Just for reference:

>>> matrix

array([[1., 0., 0., 0.],   #<- "row" index 0

       [2., 1., 0., 0.],   #<- "row" index 1

       [3., 2., 1., 0.],   #<- "row" index 2

       [4., 3., 2., 1.]])  #<- "row" index 3



      # ^      ...  ^

      # "col" 0     "col" 3

answered Nov 21 at 23:33

sacul

29.7k41640

You can add the following print statement, and the loop will explain itself at each iteration:

n=4

matrix=np.zeros((n,n))

for i in range (0,n):

    for j in range(0,i+1):

        print(f'inserting {i-j+1} into the matrix at row index {i}, columns index {j}')

        matrix[i,j]=i-j+1

When you run it, you get this output:

inserting 1 into the matrix at row index 0, columns index 0

inserting 2 into the matrix at row index 1, columns index 0

inserting 1 into the matrix at row index 1, columns index 1

...

inserting 3 into the matrix at row index 3, columns index 1

inserting 2 into the matrix at row index 3, columns index 2

inserting 1 into the matrix at row index 3, columns index 3

And your matrix is populated as before:

>>> matrix

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

       [2., 1., 0., 0.],

       [3., 2., 1., 0.],

       [4., 3., 2., 1.]])

Just for reference:

>>> matrix

array([[1., 0., 0., 0.],   #<- "row" index 0

       [2., 1., 0., 0.],   #<- "row" index 1

       [3., 2., 1., 0.],   #<- "row" index 2

       [4., 3., 2., 1.]])  #<- "row" index 3



      # ^      ...  ^

      # "col" 0     "col" 3

answered Nov 21 at 23:33

sacul

29.7k41640

answered Nov 21 at 23:33

sacul

29.7k41640

answered Nov 21 at 23:33

sacul

29.7k41640

answered Nov 21 at 23:33

sacul

29.7k41640

2

Thank you very much, this was really helpful!!
– Aylín Pérez
Nov 21 at 23:47

add a comment |

2

Thank you very much, this was really helpful!!
– Aylín Pérez
Nov 21 at 23:47

Thank you very much, this was really helpful!!
– Aylín Pérez
Nov 21 at 23:47

add a comment |

up vote
1
down vote

import numpy as np

n=4

We start by setting a 4x4 matrix with all coordinates set to null:

matrix=np.zeros((n,n))

We set new coordinate values by looping through rows and columns. First we loop through rows, from index 0 to n-1:

for i in range (0,n):

    for j in range(0,i+1):

Finally, we set the desired value for the current coordinate:

        matrix[i,j]=i-j+1

print(matrix)

answered Nov 21 at 23:46

DavidPM

18117

Thank you, really helpful!
– Aylín Pérez
Nov 21 at 23:53

add a comment |

up vote
1
down vote

import numpy as np

n=4

We start by setting a 4x4 matrix with all coordinates set to null:

matrix=np.zeros((n,n))

We set new coordinate values by looping through rows and columns. First we loop through rows, from index 0 to n-1:

for i in range (0,n):

    for j in range(0,i+1):

Finally, we set the desired value for the current coordinate:

        matrix[i,j]=i-j+1

print(matrix)

answered Nov 21 at 23:46

DavidPM

18117

Thank you, really helpful!
– Aylín Pérez
Nov 21 at 23:53

add a comment |

up vote
1
down vote

import numpy as np

n=4

We start by setting a 4x4 matrix with all coordinates set to null:

matrix=np.zeros((n,n))

We set new coordinate values by looping through rows and columns. First we loop through rows, from index 0 to n-1:

for i in range (0,n):

    for j in range(0,i+1):

Finally, we set the desired value for the current coordinate:

        matrix[i,j]=i-j+1

print(matrix)

answered Nov 21 at 23:46

DavidPM

18117

import numpy as np

n=4

We start by setting a 4x4 matrix with all coordinates set to null:

matrix=np.zeros((n,n))

We set new coordinate values by looping through rows and columns. First we loop through rows, from index 0 to n-1:

for i in range (0,n):

    for j in range(0,i+1):

Finally, we set the desired value for the current coordinate:

        matrix[i,j]=i-j+1

print(matrix)

answered Nov 21 at 23:46

DavidPM

18117

answered Nov 21 at 23:46

DavidPM

18117

answered Nov 21 at 23:46

DavidPM

18117

answered Nov 21 at 23:46

DavidPM

18117

Thank you, really helpful!
– Aylín Pérez
Nov 21 at 23:53

add a comment |

Thank you, really helpful!
– Aylín Pérez
Nov 21 at 23:53

Thank you, really helpful!
– Aylín Pérez
Nov 21 at 23:53

add a comment |

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