Fastest way to go from linear index to grid index
I'm sure this has been asked before but I'm interested in going from a position in a vector to the index in the grid version of the vector with given strides, for example, say I have the vector:
vec = {58, 94, 19, 68, 54, 77, 1, 18, 49, 20, 90, 44, 91, 89, 15, 0,
60, 18, 19, 44, 87, 5, 8, 42, 51, 55, 87, 71, 83, 68, 53, 58, 27,
17, 8, 14, 33, 58, 86, 3, 91, 66, 3, 16, 98, 84, 72, 98, 9, 30, 90,
99, 15, 0, 82, 76, 86, 58, 77, 58};
And say I have strides {5, 4, 3}, the position 35 in the vector would correspond to the index {3, 4, 2}:
vec[[35]]
4
ArrayReshape[vec, {5, 4, 3}][[3, 4, 2]]
4
How can I get this index fast and in a vectorized fashion because I will have potentially many positions to extract?
list-manipulation performance-tuning
asked 1 hour ago
b3m2a1
26.7k257154
add a comment |
I'm sure this has been asked before but I'm interested in going from a position in a vector to the index in the grid version of the vector with given strides, for example, say I have the vector:
vec = {58, 94, 19, 68, 54, 77, 1, 18, 49, 20, 90, 44, 91, 89, 15, 0,
60, 18, 19, 44, 87, 5, 8, 42, 51, 55, 87, 71, 83, 68, 53, 58, 27,
17, 8, 14, 33, 58, 86, 3, 91, 66, 3, 16, 98, 84, 72, 98, 9, 30, 90,
99, 15, 0, 82, 76, 86, 58, 77, 58};
And say I have strides {5, 4, 3}, the position 35 in the vector would correspond to the index {3, 4, 2}:
vec[[35]]
4
ArrayReshape[vec, {5, 4, 3}][[3, 4, 2]]
4
How can I get this index fast and in a vectorized fashion because I will have potentially many positions to extract?
list-manipulation performance-tuning
asked 1 hour ago
b3m2a1
26.7k257154
add a comment |
I'm sure this has been asked before but I'm interested in going from a position in a vector to the index in the grid version of the vector with given strides, for example, say I have the vector:
vec = {58, 94, 19, 68, 54, 77, 1, 18, 49, 20, 90, 44, 91, 89, 15, 0,
60, 18, 19, 44, 87, 5, 8, 42, 51, 55, 87, 71, 83, 68, 53, 58, 27,
17, 8, 14, 33, 58, 86, 3, 91, 66, 3, 16, 98, 84, 72, 98, 9, 30, 90,
99, 15, 0, 82, 76, 86, 58, 77, 58};
And say I have strides {5, 4, 3}, the position 35 in the vector would correspond to the index {3, 4, 2}:
vec[[35]]
4
ArrayReshape[vec, {5, 4, 3}][[3, 4, 2]]
4
How can I get this index fast and in a vectorized fashion because I will have potentially many positions to extract?
list-manipulation performance-tuning
asked 1 hour ago
b3m2a1
26.7k257154
I'm sure this has been asked before but I'm interested in going from a position in a vector to the index in the grid version of the vector with given strides, for example, say I have the vector:
vec = {58, 94, 19, 68, 54, 77, 1, 18, 49, 20, 90, 44, 91, 89, 15, 0,
60, 18, 19, 44, 87, 5, 8, 42, 51, 55, 87, 71, 83, 68, 53, 58, 27,
17, 8, 14, 33, 58, 86, 3, 91, 66, 3, 16, 98, 84, 72, 98, 9, 30, 90,
99, 15, 0, 82, 76, 86, 58, 77, 58};
And say I have strides {5, 4, 3}, the position 35 in the vector would correspond to the index {3, 4, 2}:
vec[[35]]
4
ArrayReshape[vec, {5, 4, 3}][[3, 4, 2]]
4
How can I get this index fast and in a vectorized fashion because I will have potentially many positions to extract?
list-manipulation performance-tuning
list-manipulation performance-tuning
asked 1 hour ago
b3m2a1
26.7k257154
asked 1 hour ago
b3m2a1
26.7k257154
asked 1 hour ago
b3m2a1
26.7k257154
asked 1 hour ago
b3m2a1
26.7k257154
asked 1 hour ago
b3m2a1
26.7k257154
26.7k257154
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
I think this is what you want:
IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1
In general:
gridIndex[n_Integer, shape_List] :=
IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1
answered 45 mins ago
swish
3,9611534
1
@C.E. You're right, it just needs dimension length specification
– swish
36 mins ago
1
This is a very nice solution, +1
– C. E.
35 mins ago
add a comment |
Here's what I came up with:
getSubindex[index_, stride_] := {
Mod[index, stride, 1],
Ceiling[index/stride]
}
getIndex[index_, strides_] :=
Reverse@FoldPairList[getSubindex, index, Reverse@strides]
This is comparable to swish's solution speed-wise:
gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000061, {3, 2, 3, 4}}
getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000052, {3, 2, 3, 4}}
answered 36 mins ago
C. E.
49.9k397202
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think this is what you want:
IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1
In general:
gridIndex[n_Integer, shape_List] :=
IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1
answered 45 mins ago
swish
3,9611534
1
@C.E. You're right, it just needs dimension length specification
– swish
36 mins ago
1
This is a very nice solution, +1
– C. E.
35 mins ago
add a comment |
I think this is what you want:
IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1
In general:
gridIndex[n_Integer, shape_List] :=
IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1
answered 45 mins ago
swish
3,9611534
1
@C.E. You're right, it just needs dimension length specification
– swish
36 mins ago
1
This is a very nice solution, +1
– C. E.
35 mins ago
add a comment |
I think this is what you want:
IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1
In general:
gridIndex[n_Integer, shape_List] :=
IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1
answered 45 mins ago
swish
3,9611534
I think this is what you want:
IntegerDigits[35 - 1, MixedRadix[{5, 4, 3}], 3] + 1
In general:
gridIndex[n_Integer, shape_List] :=
IntegerDigits[n - 1, MixedRadix[shape], Length@shape] + 1
answered 45 mins ago
swish
3,9611534
edited 33 mins ago
answered 45 mins ago
swish
3,9611534
answered 45 mins ago
swish
3,9611534
answered 45 mins ago
swish
3,9611534
3,9611534
1
@C.E. You're right, it just needs dimension length specification
– swish
36 mins ago
1
This is a very nice solution, +1
– C. E.
35 mins ago
add a comment |
1
@C.E. You're right, it just needs dimension length specification
– swish
36 mins ago
1
This is a very nice solution, +1
– C. E.
35 mins ago
1
1
@C.E. You're right, it just needs dimension length specification
– swish
36 mins ago
@C.E. You're right, it just needs dimension length specification
– swish
36 mins ago
1
1
This is a very nice solution, +1
– C. E.
35 mins ago
This is a very nice solution, +1
– C. E.
35 mins ago
add a comment |
Here's what I came up with:
getSubindex[index_, stride_] := {
Mod[index, stride, 1],
Ceiling[index/stride]
}
getIndex[index_, strides_] :=
Reverse@FoldPairList[getSubindex, index, Reverse@strides]
This is comparable to swish's solution speed-wise:
gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000061, {3, 2, 3, 4}}
getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000052, {3, 2, 3, 4}}
answered 36 mins ago
C. E.
49.9k397202
add a comment |
Here's what I came up with:
getSubindex[index_, stride_] := {
Mod[index, stride, 1],
Ceiling[index/stride]
}
getIndex[index_, strides_] :=
Reverse@FoldPairList[getSubindex, index, Reverse@strides]
This is comparable to swish's solution speed-wise:
gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000061, {3, 2, 3, 4}}
getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000052, {3, 2, 3, 4}}
answered 36 mins ago
C. E.
49.9k397202
add a comment |
Here's what I came up with:
getSubindex[index_, stride_] := {
Mod[index, stride, 1],
Ceiling[index/stride]
}
getIndex[index_, strides_] :=
Reverse@FoldPairList[getSubindex, index, Reverse@strides]
This is comparable to swish's solution speed-wise:
gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000061, {3, 2, 3, 4}}
getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000052, {3, 2, 3, 4}}
answered 36 mins ago
C. E.
49.9k397202
Here's what I came up with:
getSubindex[index_, stride_] := {
Mod[index, stride, 1],
Ceiling[index/stride]
}
getIndex[index_, strides_] :=
Reverse@FoldPairList[getSubindex, index, Reverse@strides]
This is comparable to swish's solution speed-wise:
gridIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000061, {3, 2, 3, 4}}
getIndex[1000, {3, 5, 4, 6}] // RepeatedTiming
{0.000052, {3, 2, 3, 4}}
answered 36 mins ago
C. E.
49.9k397202
edited 15 mins ago
answered 36 mins ago
C. E.
49.9k397202
answered 36 mins ago
C. E.
49.9k397202
answered 36 mins ago
C. E.
49.9k397202
49.9k397202
add a comment |
add a comment |
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