Meaning of ## #
$begingroup$
While looking for an answer to a different problem, I came across the following example:
In[1]:= Array[Plus[##] &, {2,2}]
Out[1]= {{2,3},{3,4}}
Having read through the documentation on Slot and SlotSequence, I understand the above example. If we think of it as a matrix $M_{ij}$, it takes the two indices and adds them up, $M_{ij}=i+j$. I also understand that it is equivalent to Array[Plus[#1,#2] &, {2, 2}].
I tried making the expression slightly more complicated:
In[2]:= Array[Plus[##,#] &, {2,2}]
Out[2]= {{3,4},{5,6}}
which acts as $M_{ij}=(i+j)+i$, or if we replace the last # with #2, $M_{ij}=(i+j)+j$.
Now, I am really stuck on interpreting:
In[3]:= Array[Plus[## #] &, {2,2}]
Out[3]= {{1,2},{4,8}}
What is the corresponding expression for $M_{ij}$?
array slot
asked Dec 7 '18 at 13:25
Jakub KryśJakub Kryś
634
$endgroup$
add a comment |
$begingroup$
While looking for an answer to a different problem, I came across the following example:
In[1]:= Array[Plus[##] &, {2,2}]
Out[1]= {{2,3},{3,4}}
Having read through the documentation on Slot and SlotSequence, I understand the above example. If we think of it as a matrix $M_{ij}$, it takes the two indices and adds them up, $M_{ij}=i+j$. I also understand that it is equivalent to Array[Plus[#1,#2] &, {2, 2}].
I tried making the expression slightly more complicated:
In[2]:= Array[Plus[##,#] &, {2,2}]
Out[2]= {{3,4},{5,6}}
which acts as $M_{ij}=(i+j)+i$, or if we replace the last # with #2, $M_{ij}=(i+j)+j$.
Now, I am really stuck on interpreting:
In[3]:= Array[Plus[## #] &, {2,2}]
Out[3]= {{1,2},{4,8}}
What is the corresponding expression for $M_{ij}$?
array slot
asked Dec 7 '18 at 13:25
Jakub KryśJakub Kryś
634
$endgroup$
1
$begingroup$
FullForm[## #]
$endgroup$
– Kuba♦
Dec 7 '18 at 13:30
$begingroup$
Thank you, as a beginner in Mathematica, I don't useFullFormas often as I should...
$endgroup$
– Jakub Kryś
Dec 7 '18 at 14:02
1
$begingroup$
Hmm, I think I need to find a place to use######&in my code. (It's perhaps less obvious what this is here than in the frontend...)
$endgroup$
– Brett Champion
Dec 7 '18 at 14:42
add a comment |
$begingroup$
While looking for an answer to a different problem, I came across the following example:
In[1]:= Array[Plus[##] &, {2,2}]
Out[1]= {{2,3},{3,4}}
Having read through the documentation on Slot and SlotSequence, I understand the above example. If we think of it as a matrix $M_{ij}$, it takes the two indices and adds them up, $M_{ij}=i+j$. I also understand that it is equivalent to Array[Plus[#1,#2] &, {2, 2}].
I tried making the expression slightly more complicated:
In[2]:= Array[Plus[##,#] &, {2,2}]
Out[2]= {{3,4},{5,6}}
which acts as $M_{ij}=(i+j)+i$, or if we replace the last # with #2, $M_{ij}=(i+j)+j$.
Now, I am really stuck on interpreting:
In[3]:= Array[Plus[## #] &, {2,2}]
Out[3]= {{1,2},{4,8}}
What is the corresponding expression for $M_{ij}$?
array slot
asked Dec 7 '18 at 13:25
Jakub KryśJakub Kryś
634
$endgroup$
While looking for an answer to a different problem, I came across the following example:
In[1]:= Array[Plus[##] &, {2,2}]
Out[1]= {{2,3},{3,4}}
Having read through the documentation on Slot and SlotSequence, I understand the above example. If we think of it as a matrix $M_{ij}$, it takes the two indices and adds them up, $M_{ij}=i+j$. I also understand that it is equivalent to Array[Plus[#1,#2] &, {2, 2}].
I tried making the expression slightly more complicated:
In[2]:= Array[Plus[##,#] &, {2,2}]
Out[2]= {{3,4},{5,6}}
which acts as $M_{ij}=(i+j)+i$, or if we replace the last # with #2, $M_{ij}=(i+j)+j$.
Now, I am really stuck on interpreting:
In[3]:= Array[Plus[## #] &, {2,2}]
Out[3]= {{1,2},{4,8}}
What is the corresponding expression for $M_{ij}$?
array slot
array slot
asked Dec 7 '18 at 13:25
Jakub KryśJakub Kryś
634
asked Dec 7 '18 at 13:25
Jakub KryśJakub Kryś
634
asked Dec 7 '18 at 13:25
Jakub KryśJakub Kryś
634
asked Dec 7 '18 at 13:25
Jakub KryśJakub Kryś
634
asked Dec 7 '18 at 13:25
Jakub KryśJakub Kryś
634
634
1
$begingroup$
FullForm[## #]
$endgroup$
– Kuba♦
Dec 7 '18 at 13:30
$begingroup$
Thank you, as a beginner in Mathematica, I don't useFullFormas often as I should...
$endgroup$
– Jakub Kryś
Dec 7 '18 at 14:02
1
$begingroup$
Hmm, I think I need to find a place to use######&in my code. (It's perhaps less obvious what this is here than in the frontend...)
$endgroup$
– Brett Champion
Dec 7 '18 at 14:42
add a comment |
1
$begingroup$
FullForm[## #]
$endgroup$
– Kuba♦
Dec 7 '18 at 13:30
$begingroup$
Thank you, as a beginner in Mathematica, I don't useFullFormas often as I should...
$endgroup$
– Jakub Kryś
Dec 7 '18 at 14:02
1
$begingroup$
Hmm, I think I need to find a place to use######&in my code. (It's perhaps less obvious what this is here than in the frontend...)
$endgroup$
– Brett Champion
Dec 7 '18 at 14:42
1
1
$begingroup$
FullForm[## #]$endgroup$
– Kuba♦
Dec 7 '18 at 13:30
$begingroup$
FullForm[## #]$endgroup$
– Kuba♦
Dec 7 '18 at 13:30
$begingroup$
Thank you, as a beginner in Mathematica, I don't use
FullForm as often as I should...$endgroup$
– Jakub Kryś
Dec 7 '18 at 14:02
$begingroup$
Thank you, as a beginner in Mathematica, I don't use
FullForm as often as I should...$endgroup$
– Jakub Kryś
Dec 7 '18 at 14:02
1
1
$begingroup$
Hmm, I think I need to find a place to use
######& in my code. (It's perhaps less obvious what this is here than in the frontend...)$endgroup$
– Brett Champion
Dec 7 '18 at 14:42
$begingroup$
Hmm, I think I need to find a place to use
######& in my code. (It's perhaps less obvious what this is here than in the frontend...)$endgroup$
– Brett Champion
Dec 7 '18 at 14:42
add a comment |
1 Answer
1
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votes
$begingroup$
Plus[## #] & is the same as Plus[Times[##, #1]], which always computes to the same as the simpler Times[##, #1] (because Plus[x] is just x).
Thus this computes the same as
Table[i^2 * j, {i, 2}, {j, 2}]
answered Dec 7 '18 at 13:56
SzabolcsSzabolcs
159k13435930
$endgroup$
add a comment |
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$begingroup$
Plus[## #] & is the same as Plus[Times[##, #1]], which always computes to the same as the simpler Times[##, #1] (because Plus[x] is just x).
Thus this computes the same as
Table[i^2 * j, {i, 2}, {j, 2}]
answered Dec 7 '18 at 13:56
SzabolcsSzabolcs
159k13435930
$endgroup$
add a comment |
$begingroup$
Plus[## #] & is the same as Plus[Times[##, #1]], which always computes to the same as the simpler Times[##, #1] (because Plus[x] is just x).
Thus this computes the same as
Table[i^2 * j, {i, 2}, {j, 2}]
answered Dec 7 '18 at 13:56
SzabolcsSzabolcs
159k13435930
$endgroup$
add a comment |
$begingroup$
Plus[## #] & is the same as Plus[Times[##, #1]], which always computes to the same as the simpler Times[##, #1] (because Plus[x] is just x).
Thus this computes the same as
Table[i^2 * j, {i, 2}, {j, 2}]
answered Dec 7 '18 at 13:56
SzabolcsSzabolcs
159k13435930
$endgroup$
Plus[## #] & is the same as Plus[Times[##, #1]], which always computes to the same as the simpler Times[##, #1] (because Plus[x] is just x).
Thus this computes the same as
Table[i^2 * j, {i, 2}, {j, 2}]
answered Dec 7 '18 at 13:56
SzabolcsSzabolcs
159k13435930
edited Dec 7 '18 at 22:05
answered Dec 7 '18 at 13:56
SzabolcsSzabolcs
159k13435930
answered Dec 7 '18 at 13:56
SzabolcsSzabolcs
159k13435930
answered Dec 7 '18 at 13:56
SzabolcsSzabolcs
159k13435930
159k13435930
add a comment |
add a comment |
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1
$begingroup$
FullForm[## #]$endgroup$
– Kuba♦
Dec 7 '18 at 13:30
$begingroup$
Thank you, as a beginner in Mathematica, I don't use
FullFormas often as I should...$endgroup$
– Jakub Kryś
Dec 7 '18 at 14:02
1
$begingroup$
Hmm, I think I need to find a place to use
######&in my code. (It's perhaps less obvious what this is here than in the frontend...)$endgroup$
– Brett Champion
Dec 7 '18 at 14:42