You can use a list comprehension:
>>> li= ['a','b','c','d','e','f','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
>>> [li[i:i+5] for i in range[0,len[li],5]]
[['a', 'b', 'c', 'd', 'e'], ['f', 'h', 'i', 'j', 'k'], ['l', 'm', 'n', 'o', 'p'], ['q', 'r', 's', 't', 'u'], ['v', 'w', 'x', 'y', 'z']]
Or, if you don't mind tuples, use zip:
>>> zip[*[iter[li]]*5]
[['a', 'b', 'c', 'd', 'e'], ['f', 'h', 'i', 'j', 'k'], ['l', 'm', 'n', 'o', 'p'], ['q', 'r', 's', 't', 'u'], ['v', 'w', 'x', 'y', 'z']]
Or apply list to the tuples:
>>> map[list, zip[*[iter[li]]*5]]
[['a', 'b', 'c', 'd', 'e'], ['f', 'h', 'i', 'j', 'k'], ['l', 'm', 'n', 'o', 'p'], ['q', 'r', 's', 't', 'u'], ['v', 'w', 'x', 'y', 'z']]
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Introduction to Identity Matrix :
The dictionary definition of an Identity Matrix is a square matrix
in which all the elements of the principal or main diagonal are 1’s and all other elements are zeros. In the below image, every matrix is an Identity Matrix.
In linear algebra, this is sometimes called as a Unit Matrix, of a square matrix [size = n x n] with ones on the main diagonal and zeros elsewhere. The identity matrix is denoted by “ I
“. Sometimes U or E is also used to denote an Identity Matrix.
A property of the identity matrix is that it leaves a matrix unchanged if it is multiplied by an Identity Matrix.
Examples:
Input : 2
Output : 1 0
0 1
Input : 4
Output : 1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
The explanation is simple. We need to make all
the elements of principal or main diagonal as
1 and everything else as 0.Program to print Identity Matrix :
The logic is simple. You need to the print 1 in those positions where row is equal to column of a matrix and make all other positions as 0.
Python3
def Identity[size]:
for row in range[0, size]:
for col in range[0, size]:
if [row == col]:
print["1 ", end=" "]
else:
print["0 ", end=" "]
print[]
size = 5
Identity[size]
Output:
1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1
Time complexity: O[R*C] where R and C is no of rows and column in matrix respectively
Program to check if a given square matrix is Identity Matrix :
Python3
MAX = 100;
def isIdentity[mat, N]:
for row in range[N]:
for col in range[N]:
if
[row == col and
mat[row][col] != 1]:
return False;
elif [row != col and
mat[row][col] != 0]:
return False;
return True;
N = 4;
mat = [[1, 0, 0, 0],
[0, 1, 0, 0],
[0, 0, 1, 0],
[0, 0, 0, 1]];
if [isIdentity[mat, N]]:
print["Yes "];
else:
print["No "];
Output:
Yes
Time complexity: O[N2] where N is number of rows and columns of matrix
Auxiliary Space:
O[1]