How to leave only the following strings? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How can I create a class of matrices programmatically?Manipulate excel file and plotHow do I partition a matrix?How to create a table of tables?Turn the following values into percentageHow to select the data in a given way?Reading CSV data from a streamHow to remove from the data the rows of with fixed number of elements?Padding lists for accurate plottingHow to remove the given columns and strings?
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How to leave only the following strings?
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How can I create a class of matrices programmatically?Manipulate excel file and plotHow do I partition a matrix?How to create a table of tables?Turn the following values into percentageHow to select the data in a given way?Reading CSV data from a streamHow to remove from the data the rows of with fixed number of elements?Padding lists for accurate plottingHow to remove the given columns and strings?
$begingroup$
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
asked 3 hours ago
John TaylorJohn Taylor
787211
$endgroup$
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
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
asked 3 hours ago
John TaylorJohn Taylor
787211
$endgroup$
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
list-manipulation data
asked 3 hours ago
John TaylorJohn Taylor
787211
asked 3 hours ago
John TaylorJohn Taylor
787211
asked 3 hours ago
John TaylorJohn Taylor
787211
asked 3 hours ago
John TaylorJohn Taylor
787211
asked 3 hours ago
John TaylorJohn Taylor
787211
787211
1
$begingroup$
e.g.SequenceCases[data, x_List, E, ___ :> x]
$endgroup$
– C. E.
3 hours ago
add a comment |
1
$begingroup$
e.g.SequenceCases[data, x_List, E, ___ :> x]
$endgroup$
– C. E.
3 hours ago
1
1
$begingroup$
e.g.
SequenceCases[data, x_List, E, ___ :> x]$endgroup$
– C. E.
3 hours ago
$begingroup$
e.g.
SequenceCases[data, x_List, E, ___ :> x]$endgroup$
– C. E.
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
answered 3 hours ago
sakrasakra
2,8231429
$endgroup$
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
answered 3 hours ago
C. E.C. E.
51.4k3101207
$endgroup$
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
$begingroup$
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
answered 3 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
answered 3 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
answered 3 hours ago
sakrasakra
2,8231429
$endgroup$
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
answered 3 hours ago
sakrasakra
2,8231429
answered 3 hours ago
sakrasakra
2,8231429
answered 3 hours ago
sakrasakra
2,8231429
answered 3 hours ago
sakrasakra
2,8231429
2,8231429
add a comment |
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
answered 3 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
answered 3 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
answered 3 hours ago
C. E.C. E.
51.4k3101207
$endgroup$
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
answered 3 hours ago
C. E.C. E.
51.4k3101207
edited 57 mins ago
answered 3 hours ago
C. E.C. E.
51.4k3101207
answered 3 hours ago
C. E.C. E.
51.4k3101207
answered 3 hours ago
C. E.C. E.
51.4k3101207
51.4k3101207
add a comment |
add a comment |
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$begingroup$
e.g.
SequenceCases[data, x_List, E, ___ :> x]$endgroup$
– C. E.
3 hours ago