Revisions to How to find first list element that differs from average of N previous elements by more than a given amount?

added 64 characters in body

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edited Sep 5, 2013 at 11:32

1.2k
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Here's an attempt (without select)

lst = {1, 2, 2, 1, 2, 5, 2, 4};
n = 4;
p = 1;
Flatten[If[Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p *StandardDeviation[Drop[#, -1]],
 Take[#, -1], {}] & /@ Partition[lst, n + 1, 1, 1]] // First

Alternatively with Select.

Select[Partition[lst, n + 1, 1, 1], 
    Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p*StandardDeviation[Drop[#, -1]] &] 
    // First // Last

For a more efficient implementation - see Mr Wizard's answer

Here's an attempt (without select)

lst = {1, 2, 2, 1, 2, 5, 2, 4};
n = 4;
p = 1;
Flatten[If[Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p *StandardDeviation[Drop[#, -1]],
 Take[#, -1], {}] & /@ Partition[lst, n + 1, 1, 1]] // First

Alternatively with Select.

Select[Partition[lst, n + 1, 1, 1], 
    Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p*StandardDeviation[Drop[#, -1]] &] 
    // First // Last

Here's an attempt (without select)

lst = {1, 2, 2, 1, 2, 5, 2, 4};
n = 4;
p = 1;
Flatten[If[Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p *StandardDeviation[Drop[#, -1]],
 Take[#, -1], {}] & /@ Partition[lst, n + 1, 1, 1]] // First

Alternatively with Select.

Select[Partition[lst, n + 1, 1, 1], 
    Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p*StandardDeviation[Drop[#, -1]] &] 
    // First // Last

For a more efficient implementation - see Mr Wizard's answer

Fixed select

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edited Sep 4, 2013 at 3:17

Jonie

1.2k
6
14

Here's an attempt (without select)

lst = {1, 2, 2, 1, 2, 5, 2, 4};
n = 4;
p = 1;
Flatten[If[Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p *StandardDeviation[Drop[#, -1]],
 Take[#, -1], {}] & /@ Partition[lst, n + 1, 1, 1]] // First

Alternatively with Select.

Flatten@Select[Partition[lstSelect[Partition[lst, n + 1, 1, 1], 
    Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p*StandardDeviation[Drop[#, -1]] &] 
    // First // Last

Here's an attempt (without select)

lst = {1, 2, 2, 1, 2, 5, 2, 4};
n = 4;
p = 1;
Flatten[If[Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p *StandardDeviation[Drop[#, -1]],
 Take[#, -1], {}] & /@ Partition[lst, n + 1, 1, 1]] // First

Alternatively with Select.

Flatten@Select[Partition[lst, n + 1, 1, 1], 
   Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p*StandardDeviation[Drop[#, -1]] &] // Last

Here's an attempt (without select)

lst = {1, 2, 2, 1, 2, 5, 2, 4};
n = 4;
p = 1;
Flatten[If[Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p *StandardDeviation[Drop[#, -1]],
 Take[#, -1], {}] & /@ Partition[lst, n + 1, 1, 1]] // First

Alternatively with Select.

Select[Partition[lst, n + 1, 1, 1], 
    Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p*StandardDeviation[Drop[#, -1]] &] 
    // First // Last

Added select method

Source Link

edited Sep 4, 2013 at 2:04

Jonie

1.2k
6
14

Here's an attempt (without select)

lst = {1, 2, 2, 1, 2, 5, 2, 4};
n = 4;
p = 1;
Flatten[If[Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p *StandardDeviation[Drop[#, -1]],
 Take[#, -1], {}] & /@ Partition[lst, n + 1, 1, 1]] // First

Alternatively with Select.

Flatten@Select[Partition[lst, n + 1, 1, 1], 
   Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p*StandardDeviation[Drop[#, -1]] &] // Last

Here's an attempt (without select)

lst = {1, 2, 2, 1, 2, 5, 2, 4};
n = 4;
p = 1;
Flatten[If[Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p *StandardDeviation[Drop[#, -1]],
 Take[#, -1], {}] & /@ Partition[lst, n + 1, 1, 1]] // First

Here's an attempt (without select)

lst = {1, 2, 2, 1, 2, 5, 2, 4};
n = 4;
p = 1;
Flatten[If[Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p *StandardDeviation[Drop[#, -1]],
 Take[#, -1], {}] & /@ Partition[lst, n + 1, 1, 1]] // First

Alternatively with Select.

Flatten@Select[Partition[lst, n + 1, 1, 1], 
   Abs[Take[#, -1] - Mean[Drop[#, -1]]][[1]] > p*StandardDeviation[Drop[#, -1]] &] // Last

Source Link

created Sep 4, 2013 at 1:54

Jonie

1.2k
6
14

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