Convert vector into a triangular matrix - vector

Question

I have a vector of 874! elements, which I want to turn into a triangular matrix (i.e. the top right hand corner of a square matrix).
Example Input:
1
2
3
4
5
6
7
8
9
10
Example Output:
1 2 4 7
3 5 8
6 9
10
Blanks could be filled with NAs. I'd prefer if the matrix were this way around.

Answer 1

I don't know which programming language do you want to use neither do I understand which order do you want your numbers to be stored in.
You should consider that having N elements, if you want to generate an square matrix its dimensions (n rows and columns) are given by:
N = (n*(n+1))/2
So a first approach (you should consider if your input vector has x^2/2 elements) in Python could be:
from math import sqrt
x = range(1,25+1) # This is your input vector
N = len(x)
#N = (n*(n+1))/2 # Number of elements being stored in a triangular matrix.
n = (-1.0+sqrt(1.0+8.0*N))/2.0 # Solve the equation given by the previous relation.
n = int(round(n)) # Making it integer...
while (n*(n+1))/2 < N: # ... losing precission so we should use the first n being able ...
if (n*(n+1))/2 < N: # ... to store your vector is used.
n += 1
res = [[0]*n for i in xrange(n)] # Here, we create a n*n matrix filled with zeros.
x = x[::-1] #Reverse the input so it can be consumed using pop (O(1) each extraction)
for j in xrange(n): # Fill the empty matrix following the triangular pattern and using...
if not x:
break
for i in xrange(j+1):
if not x:
break
res[i][j] = x.pop() # The elements from your input vector.
for row in res: # Let's print the result!
print(row)
The idea is to consume x filling the square matrix (res) with is values in the right order. This can easily done once you now your target matrix dimesions.