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Network questions

How to leave only the following strings?

2

$begingroup$

Consider a data having the form

data = 1,7,4,6,1,6,4,8,2,4,9,2,E,...,1,4,6,3,4,4,6,2,E,...,...

i.e., some number $n_1$ of rows followed by row $E,...$, then some number $n_2$ of rows followed by row $E,...$ and so on.

Could you please tell me how to leave only the last rows before $E,$, i.e. to obtain

subdata= 2,4,9,2,4,4,6,2,...?

list-manipulation data

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  e.g. SequenceCases[data, x_List, E, ___ :> x]
  $endgroup$
  – C. E.
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Consider a data having the form

data = 1,7,4,6,1,6,4,8,2,4,9,2,E,...,1,4,6,3,4,4,6,2,E,...,...

i.e., some number $n_1$ of rows followed by row $E,...$, then some number $n_2$ of rows followed by row $E,...$ and so on.

Could you please tell me how to leave only the last rows before $E,$, i.e. to obtain

subdata= 2,4,9,2,4,4,6,2,...?

list-manipulation data

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  e.g. SequenceCases[data, x_List, E, ___ :> x]
  $endgroup$
  – C. E.
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider a data having the form

data = 1,7,4,6,1,6,4,8,2,4,9,2,E,...,1,4,6,3,4,4,6,2,E,...,...

i.e., some number $n_1$ of rows followed by row $E,...$, then some number $n_2$ of rows followed by row $E,...$ and so on.

Could you please tell me how to leave only the last rows before $E,$, i.e. to obtain

subdata= 2,4,9,2,4,4,6,2,...?

list-manipulation data

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

$endgroup$

data = 1,7,4,6,1,6,4,8,2,4,9,2,E,...,1,4,6,3,4,4,6,2,E,...,...

subdata= 2,4,9,2,4,4,6,2,...?

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

share|improve this question

asked 3 hours ago

John TaylorJohn Taylor

787211

asked 3 hours ago

John TaylorJohn Taylor

787211

asked 3 hours ago

John TaylorJohn Taylor

787211

2 Answers
2

active

oldest

votes

2

$begingroup$

Try SequenceCases:

data = 1, 7, 4, 6, 1, 6, 4, 8, 2, 4, 9, 2, E, 1, 2, 3, 
 1, 4, 6, 3, 4, 4, 6, 2, E, 4, 5, 6
SequenceCases[data, p_, E, ___ :> p]

yields

2, 4, 9, 2, 4, 4, 6, 2

share|improve this answer

answered 2 hours ago

sakrasakra

2,8231429

$endgroup$

add a comment | 

2

$begingroup$

The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):

SequenceCases[data, x_List, E, ___ :> x]

2, 4, 9, 2, 4, 4, 6, 2

But the problem also lends itself to functional solutions, e.g.:

pairs = Partition[data, 2, 1];
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

2, 4, 9, 2, 4, 4, 6, 2

Or in one go:

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

2, 4, 9, 2, 4, 4, 6, 2

share|improve this answer

edited 48 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

$endgroup$

add a comment | 

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2

$begingroup$

Try SequenceCases:

data = 1, 7, 4, 6, 1, 6, 4, 8, 2, 4, 9, 2, E, 1, 2, 3, 
 1, 4, 6, 3, 4, 4, 6, 2, E, 4, 5, 6
SequenceCases[data, p_, E, ___ :> p]

yields

2, 4, 9, 2, 4, 4, 6, 2

share|improve this answer

answered 2 hours ago

sakrasakra

2,8231429

$endgroup$

add a comment | 

2

$begingroup$

Try SequenceCases:

data = 1, 7, 4, 6, 1, 6, 4, 8, 2, 4, 9, 2, E, 1, 2, 3, 
 1, 4, 6, 3, 4, 4, 6, 2, E, 4, 5, 6
SequenceCases[data, p_, E, ___ :> p]

yields

2, 4, 9, 2, 4, 4, 6, 2

share|improve this answer

answered 2 hours ago

sakrasakra

2,8231429

$endgroup$

add a comment | 

$begingroup$

Try SequenceCases:

data = 1, 7, 4, 6, 1, 6, 4, 8, 2, 4, 9, 2, E, 1, 2, 3, 
 1, 4, 6, 3, 4, 4, 6, 2, E, 4, 5, 6
SequenceCases[data, p_, E, ___ :> p]

yields

2, 4, 9, 2, 4, 4, 6, 2

share|improve this answer

answered 2 hours ago

sakrasakra

2,8231429

$endgroup$

data = 1, 7, 4, 6, 1, 6, 4, 8, 2, 4, 9, 2, E, 1, 2, 3, 
 1, 4, 6, 3, 4, 4, 6, 2, E, 4, 5, 6
SequenceCases[data, p_, E, ___ :> p]

2, 4, 9, 2, 4, 4, 6, 2

share|improve this answer

answered 2 hours ago

sakrasakra

2,8231429

answered 2 hours ago

sakrasakra

2,8231429

answered 2 hours ago

sakrasakra

2,8231429

2

$begingroup$

The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):

SequenceCases[data, x_List, E, ___ :> x]

2, 4, 9, 2, 4, 4, 6, 2

But the problem also lends itself to functional solutions, e.g.:

pairs = Partition[data, 2, 1];
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

2, 4, 9, 2, 4, 4, 6, 2

Or in one go:

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

2, 4, 9, 2, 4, 4, 6, 2

share|improve this answer

edited 48 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

$endgroup$

add a comment | 

2

$begingroup$

The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):

SequenceCases[data, x_List, E, ___ :> x]

2, 4, 9, 2, 4, 4, 6, 2

But the problem also lends itself to functional solutions, e.g.:

pairs = Partition[data, 2, 1];
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

2, 4, 9, 2, 4, 4, 6, 2

Or in one go:

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

2, 4, 9, 2, 4, 4, 6, 2

share|improve this answer

edited 48 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

$endgroup$

add a comment | 

$begingroup$

The most idiomatic solution to this problem is, in my opinion, pattern matching (as Sakra has also answered):

SequenceCases[data, x_List, E, ___ :> x]

2, 4, 9, 2, 4, 4, 6, 2

But the problem also lends itself to functional solutions, e.g.:

pairs = Partition[data, 2, 1];
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

2, 4, 9, 2, 4, 4, 6, 2

Or in one go:

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

2, 4, 9, 2, 4, 4, 6, 2

share|improve this answer

edited 48 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

$endgroup$

SequenceCases[data, x_List, E, ___ :> x]

pairs = Partition[data, 2, 1];
If[#[[2, 1]] == E, #[[1]], Nothing] & /@ pairs

BlockMap[If[#[[2, 1]] == E, #[[1]], Nothing] &, data, 2, 1]

share|improve this answer

edited 48 mins ago

answered 2 hours ago

C. E.C. E.

51.4k3101207

answered 2 hours ago

C. E.C. E.

51.4k3101207

answered 2 hours ago

C. E.C. E.

51.4k3101207