I have this code:
#!/usr/bin/python3
def contract(e, i, c, n):
l = len(e)
grid = [[0 for i in range(i + 1)] for x in range(l)]
for num1, row1 in enumerate(grid):
row1[0] = e[num1] #add exponents
for num2, row2 in enumerate(grid):
if 0 <= num2 < n[0]:
grid[num2][1] = c[num2]
if n[0] <= num2 < n[0] + n[1]:
grid[num2][2] = c[num2]
if n[0] + n[1] <= num2 < n[0] + n[1] + n[2]:
grid[num2][3] = c[num2]
for g in grid:
print(g)
e = [0, 1, 2, 3]
i = 3
c = [4, 5, 6, 7]
n = [1, 2, 1]
contract(e, i, c, n)
The idea of this code is that I have 2 dimensional grid that has dimensions len(e) x (i + 1). The first column contains exponents e. The rest of the columns should contain coefficients c in such a way that n determines the positions of the coefficients in the grid. For example, since n[0] = 1, column 1, row 0 in the grid contains number 4. The next element in n is 2, so the next column in grid (column 2) should contain 2 numbers, meaning numbers 5 and 6 in rows below the row that I used previously (meaning rows 1 and 2 beause row 0 is already used). n[2] = 1 so grid[3][3] = 7, etc.
I implemented this with repetitive if-statements and the code works fine, the output is as it should be:
[0, 4, 0, 0]
[1, 0, 5, 0]
[2, 0, 6, 0]
[3, 0, 0, 7]
However, I would like to make a general code that can do this for any number of coefficients and exponents. How can I convert those repetitive if statements to a single loop?
>Solution :
I would convert it into a for loop that keeps track of the sum of the elements seen so far, adjusting the corresponding element if the inequality holds for that iteration:
for num2, row2 in enumerate(grid):
total = 0
for n_idx, n_elem in enumerate(n):
if total <= num2 < total + n_elem:
grid[num2][n_idx + 1] = c[num2]
total += n_elem
I would advise against using sum() in this loop, as it recomputes the sum from scratch on each iteration, which isn’t very efficient.