Alternatives of lists vs Alternatives of strings
$begingroup$
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
asked Nov 24 '18 at 20:03
Suite401Suite401
981312
$endgroup$
add a comment |
$begingroup$
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
asked Nov 24 '18 at 20:03
Suite401Suite401
981312
$endgroup$
add a comment |
$begingroup$
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
asked Nov 24 '18 at 20:03
Suite401Suite401
981312
$endgroup$
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 Nov 24 '18 at 20:03
Suite401Suite401
981312
asked Nov 24 '18 at 20:03
Suite401Suite401
981312
asked Nov 24 '18 at 20:03
Suite401Suite401
981312
asked Nov 24 '18 at 20:03
Suite401Suite401
981312
asked Nov 24 '18 at 20:03
Suite401Suite401
981312
981312
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Use lis[[All, 1]] or First /@ list. See Part and First.
answered Nov 24 '18 at 20:06
SzabolcsSzabolcs
159k13433930
$endgroup$
add a comment |
$begingroup$
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".
answered Nov 24 '18 at 20:22
kglrkglr
179k9198410
$endgroup$
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
$begingroup$
Use lis[[All, 1]] or First /@ list. See Part and First.
answered Nov 24 '18 at 20:06
SzabolcsSzabolcs
159k13433930
$endgroup$
add a comment |
$begingroup$
Use lis[[All, 1]] or First /@ list. See Part and First.
answered Nov 24 '18 at 20:06
SzabolcsSzabolcs
159k13433930
$endgroup$
add a comment |
$begingroup$
Use lis[[All, 1]] or First /@ list. See Part and First.
answered Nov 24 '18 at 20:06
SzabolcsSzabolcs
159k13433930
$endgroup$
Use lis[[All, 1]] or First /@ list. See Part and First.
answered Nov 24 '18 at 20:06
SzabolcsSzabolcs
159k13433930
answered Nov 24 '18 at 20:06
SzabolcsSzabolcs
159k13433930
answered Nov 24 '18 at 20:06
SzabolcsSzabolcs
159k13433930
answered Nov 24 '18 at 20:06
SzabolcsSzabolcs
159k13433930
159k13433930
add a comment |
add a comment |
$begingroup$
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".
answered Nov 24 '18 at 20:22
kglrkglr
179k9198410
$endgroup$
add a comment |
$begingroup$
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".
answered Nov 24 '18 at 20:22
kglrkglr
179k9198410
$endgroup$
add a comment |
$begingroup$
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".
answered Nov 24 '18 at 20:22
kglrkglr
179k9198410
$endgroup$
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".
answered Nov 24 '18 at 20:22
kglrkglr
179k9198410
edited Nov 25 '18 at 8:20
answered Nov 24 '18 at 20:22
kglrkglr
179k9198410
answered Nov 24 '18 at 20:22
kglrkglr
179k9198410
answered Nov 24 '18 at 20:22
kglrkglr
179k9198410
179k9198410
add a comment |
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