Alternatives of lists vs Alternatives of strings
up vote
4
down vote
favorite
Obviously I am missing something obvious.
I have:
lis = {"a"}|{"b"}|{"c"}
but I want:
lis2 = "a"|"b"|"c"
Thanks as always for suggestions...
list-manipulation
add a comment |
up vote
4
down vote
favorite
Obviously I am missing something obvious.
I have:
lis = {"a"}|{"b"}|{"c"}
but I want:
lis2 = "a"|"b"|"c"
Thanks as always for suggestions...
list-manipulation
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Obviously I am missing something obvious.
I have:
lis = {"a"}|{"b"}|{"c"}
but I want:
lis2 = "a"|"b"|"c"
Thanks as always for suggestions...
list-manipulation
Obviously I am missing something obvious.
I have:
lis = {"a"}|{"b"}|{"c"}
but I want:
lis2 = "a"|"b"|"c"
Thanks as always for suggestions...
list-manipulation
list-manipulation
asked 2 days ago
Suite401
959312
959312
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Use lis[[All, 1]]
or First /@ list
. See Part
and First
.
add a comment |
up vote
6
down vote
You can also Apply (@@@)
Sequence
at level 1:
Sequence @@@ lis
"a" | "b" | "c"
Also
## & @@@ lis
"a" | "b" | "c"
Alternatively, use ReplaceAll
to replace List
with Sequence
:
lis /. List -> Sequence
"a" | "b" | "c"
Additional methods:
MapAt[Sequence &, lis, {All, 0}]
ReplacePart[lis, {_, 0} :> Sequence]
lis[[All, 0]] = Sequence; lis
all give "a" | "b" | "c"
.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Use lis[[All, 1]]
or First /@ list
. See Part
and First
.
add a comment |
up vote
6
down vote
accepted
Use lis[[All, 1]]
or First /@ list
. See Part
and First
.
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Use lis[[All, 1]]
or First /@ list
. See Part
and First
.
Use lis[[All, 1]]
or First /@ list
. See Part
and First
.
answered 2 days ago
Szabolcs
157k13430917
157k13430917
add a comment |
add a comment |
up vote
6
down vote
You can also Apply (@@@)
Sequence
at level 1:
Sequence @@@ lis
"a" | "b" | "c"
Also
## & @@@ lis
"a" | "b" | "c"
Alternatively, use ReplaceAll
to replace List
with Sequence
:
lis /. List -> Sequence
"a" | "b" | "c"
Additional methods:
MapAt[Sequence &, lis, {All, 0}]
ReplacePart[lis, {_, 0} :> Sequence]
lis[[All, 0]] = Sequence; lis
all give "a" | "b" | "c"
.
add a comment |
up vote
6
down vote
You can also Apply (@@@)
Sequence
at level 1:
Sequence @@@ lis
"a" | "b" | "c"
Also
## & @@@ lis
"a" | "b" | "c"
Alternatively, use ReplaceAll
to replace List
with Sequence
:
lis /. List -> Sequence
"a" | "b" | "c"
Additional methods:
MapAt[Sequence &, lis, {All, 0}]
ReplacePart[lis, {_, 0} :> Sequence]
lis[[All, 0]] = Sequence; lis
all give "a" | "b" | "c"
.
add a comment |
up vote
6
down vote
up vote
6
down vote
You can also Apply (@@@)
Sequence
at level 1:
Sequence @@@ lis
"a" | "b" | "c"
Also
## & @@@ lis
"a" | "b" | "c"
Alternatively, use ReplaceAll
to replace List
with Sequence
:
lis /. List -> Sequence
"a" | "b" | "c"
Additional methods:
MapAt[Sequence &, lis, {All, 0}]
ReplacePart[lis, {_, 0} :> Sequence]
lis[[All, 0]] = Sequence; lis
all give "a" | "b" | "c"
.
You can also Apply (@@@)
Sequence
at level 1:
Sequence @@@ lis
"a" | "b" | "c"
Also
## & @@@ lis
"a" | "b" | "c"
Alternatively, use ReplaceAll
to replace List
with Sequence
:
lis /. List -> Sequence
"a" | "b" | "c"
Additional methods:
MapAt[Sequence &, lis, {All, 0}]
ReplacePart[lis, {_, 0} :> Sequence]
lis[[All, 0]] = Sequence; lis
all give "a" | "b" | "c"
.
edited yesterday
answered 2 days ago
kglr
173k8194400
173k8194400
add a comment |
add a comment |
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