Select certain terms in an expression
4
$begingroup$
I would like to have a function to select terms containing q[i] terms in expressions, for example, for an expression
a q[i]+b +c q[j]+d
I would like the function to return
a q[i]+c q[j]
as result. Also at the same time, for a bare
q[i]
the same function would return
q[i]
How to realize such function? Thanks
pattern-matching filtering
asked Jan 11 at 6:34
XiaoaiXXiaoaiX
1155
$endgroup$
add a comment |
4
$begingroup$
I would like to have a function to select terms containing q[i] terms in expressions, for example, for an expression
a q[i]+b +c q[j]+d
I would like the function to return
a q[i]+c q[j]
as result. Also at the same time, for a bare
q[i]
the same function would return
q[i]
How to realize such function? Thanks
pattern-matching filtering
asked Jan 11 at 6:34
XiaoaiXXiaoaiX
1155
$endgroup$
add a comment |
4
4
4
$begingroup$
I would like to have a function to select terms containing q[i] terms in expressions, for example, for an expression
a q[i]+b +c q[j]+d
I would like the function to return
a q[i]+c q[j]
as result. Also at the same time, for a bare
q[i]
the same function would return
q[i]
How to realize such function? Thanks
pattern-matching filtering
asked Jan 11 at 6:34
XiaoaiXXiaoaiX
1155
$endgroup$
I would like to have a function to select terms containing q[i] terms in expressions, for example, for an expression
a q[i]+b +c q[j]+d
I would like the function to return
a q[i]+c q[j]
as result. Also at the same time, for a bare
q[i]
the same function would return
q[i]
How to realize such function? Thanks
pattern-matching filtering
pattern-matching filtering
asked Jan 11 at 6:34
XiaoaiXXiaoaiX
1155
asked Jan 11 at 6:34
XiaoaiXXiaoaiX
1155
edited Jan 11 at 7:21
XiaoaiX
asked Jan 11 at 6:34
XiaoaiXXiaoaiX
1155
asked Jan 11 at 6:34
XiaoaiXXiaoaiX
1155
asked Jan 11 at 6:34
XiaoaiXXiaoaiX
1155
1155
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
3
$begingroup$
expr = a q[i] + b + c q[j] + d;
f = DeleteCases[#, Except[_. _q], If[Head[#] === q, 0, 1]] &
f @ expr
a q[i] + c q[j]
f @ q[i]
q[i]
Alternatively,
f2 = Select[#, Function[x, If[Head@# === q, True, Not @ FreeQ[x, _q]]]] &;
f2 @ expr
a q[i] + c q[j]
f2 @ q[i]
q[i]
answered Jan 11 at 6:58
kglrkglr
179k9199410
$endgroup$
$begingroup$
This realize the first requirement, but for the bare $q[i]$, it returns $q$
$endgroup$
– XiaoaiX
Jan 11 at 7:24
$begingroup$
@XiaoaiX, I will update if i find a fix.
$endgroup$
– kglr
Jan 11 at 7:39
$begingroup$
thanks, from your answer, I also got confused about how the Select and Cases handle the List argument, see the question I just post here mathematica.stackexchange.com/questions/189270/…
$endgroup$
– XiaoaiX
Jan 11 at 7:46
add a comment |
3
$begingroup$
A pattern-based approach.
fn[x_. y_q + z_.] := x y + fn[z]
_fn = 0;
a q[i] + b + c q[j] + d // fn
q[i] // fn
a q[i] + c q[j]
q[i]
answered Jan 11 at 8:47
Mr.Wizard♦Mr.Wizard
231k294751042
$endgroup$
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
3
$begingroup$
expr = a q[i] + b + c q[j] + d;
f = DeleteCases[#, Except[_. _q], If[Head[#] === q, 0, 1]] &
f @ expr
a q[i] + c q[j]
f @ q[i]
q[i]
Alternatively,
f2 = Select[#, Function[x, If[Head@# === q, True, Not @ FreeQ[x, _q]]]] &;
f2 @ expr
a q[i] + c q[j]
f2 @ q[i]
q[i]
answered Jan 11 at 6:58
kglrkglr
179k9199410
$endgroup$
$begingroup$
This realize the first requirement, but for the bare $q[i]$, it returns $q$
$endgroup$
– XiaoaiX
Jan 11 at 7:24
$begingroup$
@XiaoaiX, I will update if i find a fix.
$endgroup$
– kglr
Jan 11 at 7:39
$begingroup$
thanks, from your answer, I also got confused about how the Select and Cases handle the List argument, see the question I just post here mathematica.stackexchange.com/questions/189270/…
$endgroup$
– XiaoaiX
Jan 11 at 7:46
add a comment |
3
$begingroup$
expr = a q[i] + b + c q[j] + d;
f = DeleteCases[#, Except[_. _q], If[Head[#] === q, 0, 1]] &
f @ expr
a q[i] + c q[j]
f @ q[i]
q[i]
Alternatively,
f2 = Select[#, Function[x, If[Head@# === q, True, Not @ FreeQ[x, _q]]]] &;
f2 @ expr
a q[i] + c q[j]
f2 @ q[i]
q[i]
answered Jan 11 at 6:58
kglrkglr
179k9199410
$endgroup$
$begingroup$
This realize the first requirement, but for the bare $q[i]$, it returns $q$
$endgroup$
– XiaoaiX
Jan 11 at 7:24
$begingroup$
@XiaoaiX, I will update if i find a fix.
$endgroup$
– kglr
Jan 11 at 7:39
$begingroup$
thanks, from your answer, I also got confused about how the Select and Cases handle the List argument, see the question I just post here mathematica.stackexchange.com/questions/189270/…
$endgroup$
– XiaoaiX
Jan 11 at 7:46
add a comment |
3
3
3
$begingroup$
expr = a q[i] + b + c q[j] + d;
f = DeleteCases[#, Except[_. _q], If[Head[#] === q, 0, 1]] &
f @ expr
a q[i] + c q[j]
f @ q[i]
q[i]
Alternatively,
f2 = Select[#, Function[x, If[Head@# === q, True, Not @ FreeQ[x, _q]]]] &;
f2 @ expr
a q[i] + c q[j]
f2 @ q[i]
q[i]
answered Jan 11 at 6:58
kglrkglr
179k9199410
$endgroup$
expr = a q[i] + b + c q[j] + d;
f = DeleteCases[#, Except[_. _q], If[Head[#] === q, 0, 1]] &
f @ expr
a q[i] + c q[j]
f @ q[i]
q[i]
Alternatively,
f2 = Select[#, Function[x, If[Head@# === q, True, Not @ FreeQ[x, _q]]]] &;
f2 @ expr
a q[i] + c q[j]
f2 @ q[i]
q[i]
answered Jan 11 at 6:58
kglrkglr
179k9199410
edited Jan 11 at 8:05
answered Jan 11 at 6:58
kglrkglr
179k9199410
answered Jan 11 at 6:58
kglrkglr
179k9199410
answered Jan 11 at 6:58
kglrkglr
179k9199410
179k9199410
$begingroup$
This realize the first requirement, but for the bare $q[i]$, it returns $q$
$endgroup$
– XiaoaiX
Jan 11 at 7:24
$begingroup$
@XiaoaiX, I will update if i find a fix.
$endgroup$
– kglr
Jan 11 at 7:39
$begingroup$
thanks, from your answer, I also got confused about how the Select and Cases handle the List argument, see the question I just post here mathematica.stackexchange.com/questions/189270/…
$endgroup$
– XiaoaiX
Jan 11 at 7:46
add a comment |
$begingroup$
This realize the first requirement, but for the bare $q[i]$, it returns $q$
$endgroup$
– XiaoaiX
Jan 11 at 7:24
$begingroup$
@XiaoaiX, I will update if i find a fix.
$endgroup$
– kglr
Jan 11 at 7:39
$begingroup$
thanks, from your answer, I also got confused about how the Select and Cases handle the List argument, see the question I just post here mathematica.stackexchange.com/questions/189270/…
$endgroup$
– XiaoaiX
Jan 11 at 7:46
$begingroup$
This realize the first requirement, but for the bare $q[i]$, it returns $q$
$endgroup$
– XiaoaiX
Jan 11 at 7:24
$begingroup$
This realize the first requirement, but for the bare $q[i]$, it returns $q$
$endgroup$
– XiaoaiX
Jan 11 at 7:24
$begingroup$
@XiaoaiX, I will update if i find a fix.
$endgroup$
– kglr
Jan 11 at 7:39
$begingroup$
@XiaoaiX, I will update if i find a fix.
$endgroup$
– kglr
Jan 11 at 7:39
$begingroup$
thanks, from your answer, I also got confused about how the Select and Cases handle the List argument, see the question I just post here mathematica.stackexchange.com/questions/189270/…
$endgroup$
– XiaoaiX
Jan 11 at 7:46
$begingroup$
thanks, from your answer, I also got confused about how the Select and Cases handle the List argument, see the question I just post here mathematica.stackexchange.com/questions/189270/…
$endgroup$
– XiaoaiX
Jan 11 at 7:46
add a comment |
3
$begingroup$
A pattern-based approach.
fn[x_. y_q + z_.] := x y + fn[z]
_fn = 0;
a q[i] + b + c q[j] + d // fn
q[i] // fn
a q[i] + c q[j]
q[i]
answered Jan 11 at 8:47
Mr.Wizard♦Mr.Wizard
231k294751042
$endgroup$
add a comment |
3
$begingroup$
A pattern-based approach.
fn[x_. y_q + z_.] := x y + fn[z]
_fn = 0;
a q[i] + b + c q[j] + d // fn
q[i] // fn
a q[i] + c q[j]
q[i]
answered Jan 11 at 8:47
Mr.Wizard♦Mr.Wizard
231k294751042
$endgroup$
add a comment |
3
3
3
$begingroup$
A pattern-based approach.
fn[x_. y_q + z_.] := x y + fn[z]
_fn = 0;
a q[i] + b + c q[j] + d // fn
q[i] // fn
a q[i] + c q[j]
q[i]
answered Jan 11 at 8:47
Mr.Wizard♦Mr.Wizard
231k294751042
$endgroup$
A pattern-based approach.
fn[x_. y_q + z_.] := x y + fn[z]
_fn = 0;
a q[i] + b + c q[j] + d // fn
q[i] // fn
a q[i] + c q[j]
q[i]
answered Jan 11 at 8:47
Mr.Wizard♦Mr.Wizard
231k294751042
answered Jan 11 at 8:47
Mr.Wizard♦Mr.Wizard
231k294751042
answered Jan 11 at 8:47
Mr.Wizard♦Mr.Wizard
231k294751042
answered Jan 11 at 8:47
Mr.Wizard♦Mr.Wizard
231k294751042
231k294751042
add a comment |
add a comment |
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