How do I use drop to drop the second element from all these sublists?
{{1, 2}, {2, 3}, {5, 4}}
I tried to do this:
Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
but it gave
{1}
but what I want is
{{1},{2},{5}}
list-manipulation
New contributor
add a comment |
{{1, 2}, {2, 3}, {5, 4}}
I tried to do this:
Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
but it gave
{1}
but what I want is
{{1},{2},{5}}
list-manipulation
New contributor
2
UseMap
(/@
) instead ofApply
(@@
).Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
is essentially equivalent toDrop[{1, 2}, {2, 3}, {5, 4}, {2}]
, which is clearly not what you want
– Lukas Lang
3 hours ago
I'd like to add to the answers here that the correct function for dropping a single element isDelete
, notDrop
.Drop
is the opposite ofTake
.
– Sjoerd Smit
1 hour ago
Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– Chris K
1 hour ago
add a comment |
{{1, 2}, {2, 3}, {5, 4}}
I tried to do this:
Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
but it gave
{1}
but what I want is
{{1},{2},{5}}
list-manipulation
New contributor
{{1, 2}, {2, 3}, {5, 4}}
I tried to do this:
Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
but it gave
{1}
but what I want is
{{1},{2},{5}}
list-manipulation
list-manipulation
New contributor
New contributor
edited 1 hour ago
Glorfindel
1971211
1971211
New contributor
asked 3 hours ago
user62264user62264
111
111
New contributor
New contributor
2
UseMap
(/@
) instead ofApply
(@@
).Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
is essentially equivalent toDrop[{1, 2}, {2, 3}, {5, 4}, {2}]
, which is clearly not what you want
– Lukas Lang
3 hours ago
I'd like to add to the answers here that the correct function for dropping a single element isDelete
, notDrop
.Drop
is the opposite ofTake
.
– Sjoerd Smit
1 hour ago
Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– Chris K
1 hour ago
add a comment |
2
UseMap
(/@
) instead ofApply
(@@
).Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
is essentially equivalent toDrop[{1, 2}, {2, 3}, {5, 4}, {2}]
, which is clearly not what you want
– Lukas Lang
3 hours ago
I'd like to add to the answers here that the correct function for dropping a single element isDelete
, notDrop
.Drop
is the opposite ofTake
.
– Sjoerd Smit
1 hour ago
Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– Chris K
1 hour ago
2
2
Use
Map
(/@
) instead of Apply
(@@
). Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
is essentially equivalent to Drop[{1, 2}, {2, 3}, {5, 4}, {2}]
, which is clearly not what you want– Lukas Lang
3 hours ago
Use
Map
(/@
) instead of Apply
(@@
). Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
is essentially equivalent to Drop[{1, 2}, {2, 3}, {5, 4}, {2}]
, which is clearly not what you want– Lukas Lang
3 hours ago
I'd like to add to the answers here that the correct function for dropping a single element is
Delete
, not Drop
. Drop
is the opposite of Take
.– Sjoerd Smit
1 hour ago
I'd like to add to the answers here that the correct function for dropping a single element is
Delete
, not Drop
. Drop
is the opposite of Take
.– Sjoerd Smit
1 hour ago
Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– Chris K
1 hour ago
Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– Chris K
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
lst = {{1, 2}, {2, 3}, {5, 4}};
Drop[lst, None, {2}]
{{1}, {2}, {5}}
add a comment |
Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}}
,
res=list[[All, 1]]
If you need the brackets around each element, Partition[res, 1]
If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.
Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}
add a comment |
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2 Answers
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2 Answers
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active
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lst = {{1, 2}, {2, 3}, {5, 4}};
Drop[lst, None, {2}]
{{1}, {2}, {5}}
add a comment |
lst = {{1, 2}, {2, 3}, {5, 4}};
Drop[lst, None, {2}]
{{1}, {2}, {5}}
add a comment |
lst = {{1, 2}, {2, 3}, {5, 4}};
Drop[lst, None, {2}]
{{1}, {2}, {5}}
lst = {{1, 2}, {2, 3}, {5, 4}};
Drop[lst, None, {2}]
{{1}, {2}, {5}}
answered 3 hours ago
kglrkglr
177k9198408
177k9198408
add a comment |
add a comment |
Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}}
,
res=list[[All, 1]]
If you need the brackets around each element, Partition[res, 1]
If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.
Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}
add a comment |
Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}}
,
res=list[[All, 1]]
If you need the brackets around each element, Partition[res, 1]
If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.
Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}
add a comment |
Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}}
,
res=list[[All, 1]]
If you need the brackets around each element, Partition[res, 1]
If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.
Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}
Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}}
,
res=list[[All, 1]]
If you need the brackets around each element, Partition[res, 1]
If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.
Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}
answered 3 hours ago
TitusTitus
605417
605417
add a comment |
add a comment |
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user62264 is a new contributor. Be nice, and check out our Code of Conduct.
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2
Use
Map
(/@
) instead ofApply
(@@
).Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}
is essentially equivalent toDrop[{1, 2}, {2, 3}, {5, 4}, {2}]
, which is clearly not what you want– Lukas Lang
3 hours ago
I'd like to add to the answers here that the correct function for dropping a single element is
Delete
, notDrop
.Drop
is the opposite ofTake
.– Sjoerd Smit
1 hour ago
Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– Chris K
1 hour ago