hiyume - 4 months ago 5x

Python Question

Say I have a list,

`A = range(1, 6) = [1, 2, 3, 4, 5]`

`B`

`i`

`j`

`i`

`j`

`B[j] = sum(A[j:i+1] or A[i:j+1])`

`j`

`i`

Examples for indices 0 and 2:

`B[0] = [1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5]`

= [1, 3, 6, 10, 15]

B[2] = [1+2+3, 2+3, 3, 3+4, 3+4+5]

= [6, 5, 3, 7, 12]

======

Current code (works) is two

`for`

`reduce`

`A = range(1,6)`

n = len(A)

B = []

for j in xrange(n):

b = []

for i in xrange(n):

if j <= i:

b.append(sum(A[j:i+1]))

else:

b.append(sum(A[i:j+1]))

B.append(b)

for b in B:

print b

Minor context: possibly part of my solution to project euler 82

Answer

You end up recalculating the sums many times. Instead create them once and look them up for each element of `b`

:

```
A = range(1,6)
n = len(A)
mapping = {}
for i in xrange(n):
for j in xrange(i,n):
mapping[i,j] = sum(A[i:j+1])
B = []
for j in xrange(n):
b = []
for i in xrange(n):
if j <= i:
b.append(mapping[j,i])
else:
b.append(mapping[i,j])
B.append(b)
```

you could eliminate the need to check `j<=i`

if you just make the mapping work for both `[i,j]`

or `[j,i]`

:

```
mapping = {}
A = range(1,6)
n = len(A)
for i in xrange(n):
for j in xrange(i,n):
mapping[i,j] = sum(A[i:j+1])
mapping[j,i] = mapping[i,j] #for both ways
B = [[mapping[i,j] for i in xrange(n)] for j in xrange(n)]
```

Although notice that this means that every `B[x][y]`

will directly coordinate to `mapping[x,y]`

so you may just want to use the mapping by itself.

Source (Stackoverflow)

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