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How to use 1D SparseArray instead of Table?

hwoarang Polar
Posted 10 years ago

Hello, everyone.

I want to ask you very very simple thing, but it looks strange to me at least. I have to use 1 dim SparseArray function instead of Table. But I don't know why the following example is not possible.

\[Chi]square = SparseArray[1 -> 3, 4]
Position[\[Chi]square, 3]

When Table is used, then it works fine (even the depth of array and table are same). Can you tell me what is reason for this ? Thank you in advance.
POSTED BY: hwoarang Polar
11 Replies
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Examples of atomic objects include SparseArray, Graph, (Boundary)MeshRegion, Association, Dispatch and a few more obscure/undocumented raw types. An exhaustive list is unlikely to be very useful.
POSTED BY: Ilian Gachevski

Maybe not useful, but still interesting.

POSTED BY: Frank Kampas

Being a physicist, I can't resist asking, "What (specifically) are the Atoms in Mathematica?" AtomQ will tell me if something is Atomic. However, the documentation lists Symbol, String, Integer, Real, Rational and Complex as Atomic Objects but that clearly is not complete,

POSTED BY: Frank Kampas
David Keith
David Keith
Posted 10 years ago

And could splitting SparseArrays be a new source of clean energy?
POSTED BY: David Keith

The reason Position doesn't work is that SparseArray objects are atomic.
POSTED BY: Ilian Gachevski 
In[4]:= Position[Normal @ \[Chi]square, 3]

Out[4]= {{1}}
POSTED BY: Frank Kampas

I was trying to point out that when you construct a SparseArray, the left-hand side of the rule is the position and the right-hand side is the value you are putting in that position. SparseArray[1->3,4] has a 3 in position 1. I admit it is counter-intuitive.
POSTED BY: Frank Kampas
hwoarang Polar
hwoarang Polar
Posted 10 years ago

Thank you,Frank for the reply.

But the strange thing is that when I use SparseArray, Position command cannot find the position of the element that I specified like below. Did I misunderstand something ?

In[125]:= \[Chi]square = SparseArray[1 -> 3, 4];
Position[\[Chi]square, 3]

Out[126]= {}
POSTED BY: hwoarang Polar
hwoarang Polar
hwoarang Polar
Posted 10 years ago

Ah, sorry. I found it.

When I use 'Normal' command, Position can read it.

In[146]:= \[Chi]square = Normal[SparseArray[1 -> 3, 4]]
Position[\[Chi]square, 3]

Out[146]= {3, 0, 0, 0}

Out[147]= {{1}}

Thank you and have a good day!
POSTED BY: hwoarang Polar
hwoarang Polar
hwoarang Polar
Posted 10 years ago

Thank you very much for the reply, Frank.

But I mean that why Position command is not working with SparseArray. This gives me blank. When I use data with parenthesis as follows, it gives me the position of the element. Can you give me a hint ?

In[98]:= \[Chi]square = {3, 1, 5, 4}

Out[98]= {3, 1, 5, 4}

In[99]:= Position[\[Chi]square, 1]

Out[99]= {{2}}
POSTED BY: hwoarang Polar 
In[1]:= \[Chi]square = SparseArray[1 -> 3, 4]

Out[1]= SparseArray[< 1 >, {4}]

In[2]:= Normal @ \[Chi]square

Out[2]= {3, 0, 0, 0}
POSTED BY: Frank Kampas
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