Store unevaluted function in list mathematica - wolfram-mathematica

Example:
list:={ Plus[1,1], Times[2,3] }
When looking at list, I get
{2,6}
I want to keep them unevaluated (as above) so that list returns
{ Plus[1,1], Times[2,3] }
Later I want to evaluate the functions in list sequence to get
{2,6}
The number of unevaluated functions in list is not known beforehand. Besides Plus, user defined functions like f[x_] may be stored in list
I hope the example is clear.
What is the best way to do this?

The best way is to store them in Hold, not List, like so:
In[255]:= f[x_] := x^2;
lh = Hold[Plus[1, 1], Times[2, 3], f[2]]
Out[256]= Hold[1 + 1, 2 3, f[2]]
In this way, you have full control over them. At some point, you may call ReleaseHold to evaluate them:
In[258]:= ReleaseHold#lh
Out[258]= Sequence[2, 6, 4]
If you want the results in a list rather than Sequence, you may use just List##lh instead. If you need to evaluate a specific one, simply use Part to extract it:
In[261]:= lh[[2]]
Out[261]= 6
If you insist on your construction, here is a way:
In[263]:= l:={Plus[1,1],Times[2,3],f[2]};
Hold[l]/.OwnValues[l]
Out[264]= Hold[{1+1,2 3,f[2]}]
EDIT
In case you have some functions/symbols with UpValues which can evaluate even inside Hold, you may want to use HoldComplete in place of Hold.
EDIT2
As pointed by #Mr.Wizard in another answer, sometimes you may find it more convenient to have Hold wrapped around individual items in your sequence. My comment here is that the usefulness of both forms is amplified once we realize that it is very easy to transform one into another and back. The following function will split the sequence inside Hold into a list of held items:
splitHeldSequence[Hold[seq___], f_: Hold] := List ## Map[f, Hold[seq]]
for example,
In[274]:= splitHeldSequence[Hold[1 + 1, 2 + 2]]
Out[274]= {Hold[1 + 1], Hold[2 + 2]}
grouping them back into a single Hold is even easier - just Apply Join:
In[275]:= Join ## {Hold[1 + 1], Hold[2 + 2]}
Out[275]= Hold[1 + 1, 2 + 2]
The two different forms are useful in diferrent circumstances. You can easily use things such as Union, Select, Cases on a list of held items without thinking much about evaluation. Once finished, you can combine them back into a single Hold, for example, to feed as unevaluated sequence of arguments to some function.
EDIT 3
Per request of #ndroock1, here is a specific example. The setup:
l = {1, 1, 1, 2, 4, 8, 3, 9, 27}
S[n_] := Module[{}, l[[n]] = l[[n]] + 1; l]
Z[n_] := Module[{}, l[[n]] = 0; l]
placing functions in Hold:
In[43]:= held = Hold[Z[1], S[1]]
Out[43]= Hold[Z[1], S[1]]
Here is how the exec function may look:
exec[n_] := MapAt[Evaluate, held, n]
Now,
In[46]:= {exec[1], exec[2]}
Out[46]= {Hold[{0, 1, 1, 2, 4, 8, 3, 9, 27}, S[1]], Hold[Z[1], {1, 1, 1, 2, 4, 8, 3, 9, 27}]}
Note that the original variable held remains unchanged, since we operate on the copy. Note also that the original setup contains mutable state (l), which is not very idiomatic in Mathematica. In particular, the order of evaluations matter:
In[61]:= Reverse[{exec[2], exec[1]}]
Out[61]= {Hold[{0, 1, 1, 2, 4, 8, 3, 9, 27}, S[1]], Hold[Z[1], {2, 1, 1, 2, 4, 8, 3, 9, 27}]}
Whether or not this is desired depends on the specific needs, I just wanted to point this out. Also, while the exec above is implemented according to the requested spec, it implicitly depends on a global variable l, which I consider a bad practice.
An alternative way to store functions suggested by #Mr.Wizard can be achieved e.g. like
In[63]:= listOfHeld = splitHeldSequence[held]
Out[63]= {Hold[Z1], Hold[S1]}
and here
In[64]:= execAlt[n_] := MapAt[ReleaseHold, listOfHeld, n]
In[70]:= l = {1, 1, 1, 2, 4, 8, 3, 9, 27} ;
{execAlt[1], execAlt[2]}
Out[71]= {{{0, 1, 1, 2, 4, 8, 3, 9, 27}, Hold[S[1]]}, {Hold[Z[1]], {1, 1, 1, 2, 4, 8, 3, 9, 27}}}
The same comments about mutability and dependence on a global variable go here as well. This last form is also more suited to query the function type:
getType[n_, lh_] := lh[[n]] /. {Hold[_Z] :> zType, Hold[_S] :> sType, _ :> unknownType}
for example:
In[172]:= getType[#, listOfHeld] & /# {1, 2}
Out[172]= {zType, sType}

The first thing that spings to mind is to not use List but rather use something like this:
SetAttributes[lst, HoldAll];
heldL=lst[Plus[1, 1], Times[2, 3]]
There will surely be lots of more erudite suggestions though!

You can also use Hold on every element that you want held:
a = {Hold[2 + 2], Hold[2*3]}
You can use HoldForm on either the elements or the list, if you want the appearance of the list without Hold visible:
b = {HoldForm[2 + 2], HoldForm[2*3]}
c = HoldForm#{2 + 2, 2*3}
{2 + 2, 2 * 3}
And you can recover the evaluated form with ReleaseHold:
a // ReleaseHold
b // ReleaseHold
c // ReleaseHold
Out[8]= {4, 6}
Out[9]= {4, 6}
Out[10]= {4, 6}
The form Hold[2+2, 2*3] or that of a and b above are good because you can easily add terms with e.g. Append. For b type is it logically:
Append[b, HoldForm[8/4]]
For Hold[2+2, 2*3]:
Hold[2+2, 2*3] ~Join~ Hold[8/4]

Another way:
lh = Function[u, Hold#u, {HoldAll, Listable}];
k = lh#{2 + 2, Sin[Pi]}
(*
->{Hold[2 + 2], Hold[Sin[\[Pi]]]}
*)
ReleaseHold#First#k
(*
-> 4
*)

Related

Wolfram Language, RandomChoice recursion

a = RandomChoice[{a,2}]&
a[]
There are other ways to achieve this example, but I wish to do more complicated things similar to this using this method.
Can I get this to continue until there are no as left to resolve, without producing a stack overflow by trying to resolve {a,2} before making the choice? Instead making the choice and resolving only the symbol chosen.
Here is a way to have RandomChoice evaluate a function only when selected:
g := (Print["evaluate g"]; 42);
f = ( If[TrueQ[#], g, #] &#RandomChoice[{True, 1, 2, 3, 4}]) &
Table[f[], {10}]
this prints "evaluate g" just when randomly selected and outputs eg.
(* {2, 42, 3, 1, 3, 2, 4, 42, 2, 4} *)
This is another way, maybe a bit cleaner:
f = Unevaluated[{g, 1, 2, 3, 4}][[RandomInteger[{1, 5}]]] &
this works fine recursively:
a = Unevaluated[{a[], 2}][[RandomInteger[{1, 2}]]] &
Though as i said in comment it simply returns 2 every time since it recurses until 2 is chosen.
a[] (* 2 *)
I do not understand the entirety of the question and my guess is there is a better way to accomplish what you want.
Your request seems slightly contradictory: when the random choice selects a you want it to recurse, but if it chooses a number you still want to continue until there are no a's left.
The following code does this, but the recursion is technically unnecessary. Perhaps in your application it will be applicable.
The first part shows how you can recurse selections of a by using Hold :-
Clear[a]
list = {1, 2, a, 1, a, 3, a, 1, 4, 5};
heldlist = Hold /# list;
a := Module[{z}, z = RandomChoice[heldlist];
Print["for information, choice was ", z];
If[MatchQ[z, Hold[_Symbol]],
heldlist = DeleteCases[heldlist, z, 1, 1];
If[MemberQ[heldlist, Hold[_Symbol]], ReleaseHold[z]]]]
a
On this occasion, calling a recurses once, then picks a 4 and stops, as expected.
for information, choice was Hold[a]
for information, choice was Hold[4]
To make the process continue until there are no a's While can be used. There is recursion going on here too but While keeps the process going when a number is picked.
While[MemberQ[heldlist, Hold[_Symbol]], a]
for information, choice was Hold[a]
for information, choice was Hold[1]
for information, choice was Hold[a]
These are the remaining items in the list :-
ReleaseHold /# heldlist
{1, 2, 1, 3, 1, 4, 5}

Deriving Mean of Values within a Tensor

I have a 20000 x 185 x 5 tensor, which looks like
{{{a1_1,a2_1,a3_1,a4_1,a5_1},{b1_1,b2_1,b3_1,b4_1,b5_1}...
(continue for 185 times)}
{{a1_2,a2_2,a3_2,a4_2,a5_2},{b1_2,b2_2,b3_2,b4_2,b5_2}...
...
...
...
{{a1_20000,a2_20000,a3_20000,a4_20000,a5_20000},
{b1_20000,b2_20000,b3_20000,b4_20000,b5_20000}... }}
The 20000 represents iteration number, the 185 represents individuals, and each individual has 5 attributes. I need to construct a 185 x 5 matrix that stores the mean value for each individual's 5 attributes, averaged across the 20000 iterations.
Not sure what the best way to do this is. I know Mean[ ] works on matrices, but with a Tensor, the derived values might not be what I need. Also, Mathematica ran out of memory if I tried to do Mean[tensor]. Please provide some help or advice. Thank you.
When in doubt, drop the size of the dimensions. (You can still keep them distinct to easily see where things end up.)
(* In[1]:= *) data = Array[a, {4, 3, 2}]
(* Out[1]= *) {{{a[1, 1, 1], a[1, 1, 2]}, {a[1, 2, 1],
a[1, 2, 2]}, {a[1, 3, 1], a[1, 3, 2]}}, {{a[2, 1, 1],
a[2, 1, 2]}, {a[2, 2, 1], a[2, 2, 2]}, {a[2, 3, 1],
a[2, 3, 2]}}, {{a[3, 1, 1], a[3, 1, 2]}, {a[3, 2, 1],
a[3, 2, 2]}, {a[3, 3, 1], a[3, 3, 2]}}, {{a[4, 1, 1],
a[4, 1, 2]}, {a[4, 2, 1], a[4, 2, 2]}, {a[4, 3, 1], a[4, 3, 2]}}}
(* In[2]:= *) Dimensions[data]
(* Out[2]= *) {4, 3, 2}
(* In[3]:= *) means = Mean[data]
(* Out[3]= *) {
{1/4 (a[1, 1, 1] + a[2, 1, 1] + a[3, 1, 1] + a[4, 1, 1]),
1/4 (a[1, 1, 2] + a[2, 1, 2] + a[3, 1, 2] + a[4, 1, 2])},
{1/4 (a[1, 2, 1] + a[2, 2, 1] + a[3, 2, 1] + a[4, 2, 1]),
1/4 (a[1, 2, 2] + a[2, 2, 2] + a[3, 2, 2] + a[4, 2, 2])},
{1/4 (a[1, 3, 1] + a[2, 3, 1] + a[3, 3, 1] + a[4, 3, 1]),
1/4 (a[1, 3, 2] + a[2, 3, 2] + a[3, 3, 2] + a[4, 3, 2])}
}
(* In[4]:= *) Dimensions[means]
(* Out[4]= *) {3, 2}
Mathematica ran out of memory if I tried to do Mean[tensor]
This is probably because intermediate results are larger than the final result. This is likely if the elements are not type Real or Integer. Example:
a = Tuples[{x, Sqrt[y], z^x, q/2, Mod[r, 1], Sin[s]}, {2, 4}];
{MemoryInUse[], MaxMemoryUsed[]}
b = Mean[a];
{MemoryInUse[], MaxMemoryUsed[]}
{109125576, 124244808}
{269465456, 376960648}
If they are, and are in packed array form, perhaps the elements are such that the array in unpacked during processing.
Here is an example where the tensor is a packed array of small numbers, and unpacking does not occur.
a = RandomReal[99, {20000, 185, 5}];
PackedArrayQ[a]
{MemoryInUse[], MaxMemoryUsed[]}
b = Mean[a];
{MemoryInUse[], MaxMemoryUsed[]}
True
{163012808, 163016952}
{163018944, 163026688}
Here is the same size of tensor with very large numbers.
a = RandomReal[$MaxMachineNumber, {20000, 185, 5}];
Developer`PackedArrayQ[a]
{MemoryInUse[], MaxMemoryUsed[]}
b = Mean[a];
{MemoryInUse[], MaxMemoryUsed[]}
True
{163010680, 458982088}
{163122608, 786958080}
To elaborate a little on the other answers, there is no reason to expect Mathematica functions to operate materially differently on tensors than matrices because Mathemetica considers them both to be nested Lists, that are just of different nesting depth. How functions behave with lists depends on whether they're Listable, which you can check using Attributes[f], where fis the function you are interested in.
Your data list's dimensionality isn't actually that big in the scheme of things. Without seeing your actual data it is hard to be sure, but I suspect the reason you are running out of memory is that some of your data is non-numerical.
I don't know what you're doing incorrectly (your code will help). But Mean[] already works as you want it to.
a = RandomReal[1, {20000, 185, 5}];
b = Mean#a;
Dimensions#b
Out[1]= {185, 5}
You can even check that this is correct:
{Max#b, Min#b}
Out[2]={0.506445, 0.494061}
which is the expected value of the mean given that RandomReal uses a uniform distribution by default.
Assume you have the following data :
a = Table[RandomInteger[100], {i, 20000}, {j, 185}, {k, 5}];
In a straightforward manner You can find a table which stores the means of a[[1,j,k]],a[[2,j,k]],...a[[20000,j,k]]:
c = Table[Sum[a[[i, j, k]], {i, Length[a]}], {j, 185}, {k, 5}]/
Length[a] // N; // Timing
{37.487, Null}
or simply :
d = Total[a]/Length[a] // N; // Timing
{0.702, Null}
The second way is about 50 times faster.
c == d
True
To extend on Brett's answer a bit, when you call Mean on a n-dimensional tensor then it averages over the first index and returns an n-1 dimensional tensor:
a = RandomReal[1, {a1, a2, a3, ... an}];
Dimensions[a] (* This would have n entries in it *)
b = Mean[a];
Dimensions[b] (* Has n-1 entries, where averaging was done over the first index *)
In the more general case where you may wish to average over the i-th argument, you would have to transpose the data around first. For example, say you want to average the 3nd of 5 dimensions. You would need the 3rd element first, followed by the 1st, 2nd, 4th, 5th.
a = RandomReal[1, {5, 10, 2, 40, 10}];
b = Transpose[a, {2, 3, 4, 1, 5}];
c = Mean[b]; (* Now of dimensions {5, 10, 40, 10} *)
In other words, you would make a call to Transpose where you placed the i-th index as the first tensor index and moved everything before it ahead one. Anything that comes after the i-th index stays the same.
This tends to come in handy when your data comes in odd formats where the first index may not always represent different realizations of a data sample. I've had this come up, for example, when I had to do time averaging of large wind data sets where the time series came third (!) in terms of the tensor representation that was available.
You could imagine the generalizedTenorMean would look something like this then:
Clear[generalizedTensorMean];
generalizedTensorMean[A_, i_] :=
Module[{n = Length#Dimensions#A, ordering},
ordering =
Join[Table[x, {x, 2, i}], {1}, Table[x, {x, i + 1, n}]];
Mean#Transpose[A, ordering]]
This reduces to the plain-old-mean when i == 1. Try it out:
A = RandomReal[1, {2, 4, 6, 8, 10, 12, 14}];
Dimensions#A (* {2, 4, 6, 8, 10, 12, 14} *)
Dimensions#generalizedTensorMean[A, 1] (* {4, 6, 8, 10, 12, 14} *)
Dimensions#generalizedTensorMean[A, 7] (* {2, 4, 6, 8, 10, 12} *)
On a side note, I'm surprised that Mathematica doesn't support this by default. You don't always want to average over the first level of a list.

Generate a list in Mathematica with a conditional tested for each element

Suppose we want to generate a list of primes p for which p + 2 is also prime.
A quick solution is to generate a complete list of the first n primes and use the Select function to return the elements which meet the condition.
Select[Table[Prime[k], {k, n}], PrimeQ[# + 2] &]
However, this is inefficient as it loads a large list into the memory before returning the filtered list. A For loop with Sow/Reap (or l = {}; AppendTo[l, k]) solves the memory issue, but it is far from elegant and is cumbersome to implement a number of times in a Mathematica script.
Reap[
For[k = 1, k <= n, k++,
p = Prime[k];
If[PrimeQ[p + 2], Sow[p]]
]
][[-1, 1]]
An ideal solution would be a built-in function which allows an option similar to this.
Table[Prime[k], {k, n}, AddIf -> PrimeQ[# + 2] &]
I will interpret this more as a question about automation and software engineering rather than about the specific problem at hand, and given a large number of solutions posted already. Reap and Sow are good means (possibly, the best in the symbolic setting) to collect intermediate results. Let us just make it general, to avoid code duplication.
What we need is to write a higher-order function. I will not do anything radically new, but will simply package your solution to make it more generally applicable:
Clear[tableGen];
tableGen[f_, iter : {i_Symbol, __}, addif : Except[_List] : (True &)] :=
Module[{sowTag},
If[# === {}, #, First##] &#
Last#Reap[Do[If[addif[#], Sow[#,sowTag]] &[f[i]], iter],sowTag]];
The advantages of using Do over For are that the loop variable is localized dynamically (so, no global modifications for it outside the scope of Do), and also the iterator syntax of Do is closer to that of Table (Do is also slightly faster).
Now, here is the usage
In[56]:= tableGen[Prime, {i, 10}, PrimeQ[# + 2] &]
Out[56]= {3, 5, 11, 17, 29}
In[57]:= tableGen[Prime, {i, 3, 10}, PrimeQ[# + 1] &]
Out[57]= {}
In[58]:= tableGen[Prime, {i, 10}]
Out[58]= {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
EDIT
This version is closer to the syntax you mentioned (it takes an expression rather than a function):
ClearAll[tableGenAlt];
SetAttributes[tableGenAlt, HoldAll];
tableGenAlt[expr_, iter_List, addif : Except[_List] : (True &)] :=
Module[{sowTag},
If[# === {}, #, First##] &#
Last#Reap[Do[If[addif[#], Sow[#,sowTag]] &[expr], iter],sowTag]];
It has an added advantage that you may even have iterator symbols defined globally, since they are passed unevaluated and dynamically localized. Examples of use:
In[65]:= tableGenAlt[Prime[i], {i, 10}, PrimeQ[# + 2] &]
Out[65]= {3, 5, 11, 17, 29}
In[68]:= tableGenAlt[Prime[i], {i, 10}]
Out[68]= {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
Note that since the syntax is different now, we had to use the Hold-attribute to prevent the passed expression expr from premature evaluation.
EDIT 2
Per #Simon's request, here is the generalization for many dimensions:
ClearAll[tableGenAltMD];
SetAttributes[tableGenAltMD, HoldAll];
tableGenAltMD[expr_, iter__List, addif : Except[_List] : (True &)] :=
Module[{indices, indexedRes, sowTag},
SetDelayed ## Prepend[Thread[Map[Take[#, 1] &, List ## Hold ### Hold[iter]],
Hold], indices];
indexedRes =
If[# === {}, #, First##] &#
Last#Reap[Do[If[addif[#], Sow[{#, indices},sowTag]] &[expr], iter],sowTag];
Map[
First,
SplitBy[indexedRes ,
Table[With[{i = i}, Function[Slot[1][[2, i]]]], {i,Length[Hold[iter]] - 1}]],
{-3}]];
It is considerably less trivial, since I had to Sow the indices together with the added values, and then split the resulting flat list according to the indices. Here is an example of use:
{i, j, k} = {1, 2, 3};
tableGenAltMD[i + j + k, {i, 1, 5}, {j, 1, 3}, {k, 1, 2}, # < 7 &]
{{{3, 4}, {4, 5}, {5, 6}}, {{4, 5}, {5, 6}, {6}}, {{5, 6}, {6}}, {{6}}}
I assigned the values to i,j,k iterator variables to illustrate that this function does localize the iterator variables and is insensitive to possible global values for them. To check the result, we may use Table and then delete the elements not satisfying the condition:
In[126]:=
DeleteCases[Table[i + j + k, {i, 1, 5}, {j, 1, 3}, {k, 1, 2}],
x_Integer /; x >= 7, Infinity] //. {} :> Sequence[]
Out[126]= {{{3, 4}, {4, 5}, {5, 6}}, {{4, 5}, {5, 6}, {6}}, {{5, 6}, {6}}, {{6}}}
Note that I did not do extensive checks so the current version may contain bugs and needs some more testing.
EDIT 3 - BUG FIX
Note the important bug-fix: in all functions, I now use Sow with a custom unique tag, and Reap as well. Without this change, the functions would not work properly when expression they evaluate also uses Sow. This is a general situation with Reap-Sow, and resembles that for exceptions (Throw-Catch).
EDIT 4 - SyntaxInformation
Since this is such a potentially useful function, it is nice to make it behave more like a built-in function. First we add syntax highlighting and basic argument checking through
SyntaxInformation[tableGenAltMD] = {"ArgumentsPattern" -> {_, {_, _, _., _.}.., _.},
"LocalVariables" -> {"Table", {2, -2}}};
Then, adding a usage message allows the menu item "Make Template" (Shift+Ctrl+k) to work:
tableGenAltMD::usage = "tableGenAltMD[expr,{i,imax},addif] will generate \
a list of values expr when i runs from 1 to imax, \
only including elements if addif[expr] returns true.
The default of addiff is True&."
A more complete and formatted usage message can be found in this gist.
I think the Reap/Sow approach is likely to be most efficient in terms of memory usage. Some alternatives might be:
DeleteCases[(With[{p=Prime[#]},If[PrimeQ[p+2],p,{}] ] ) & /# Range[K]),_List]
Or (this one might need some sort of DeleteCases to eliminate Null results):
FoldList[[(With[{p=Prime[#2]},If[PrimeQ[p+2],p] ] )& ,1.,Range[2,K] ]
Both hold a big list of integers 1 to K in memory, but the Primes are scoped inside the With[] construct.
Yes, this is another answer. Another alternative that includes the flavour of the Reap/Sow approach and the FoldList approach would be to use Scan.
result = {1};
Scan[With[{p=Prime[#]},If[PrimeQ[p+2],result={result,p}]]&,Range[2,K] ];
Flatten[result]
Again, this involves a long list of integers, but the intermediate Prime results are not stored because they are in the local scope of With. Because p is a constant in the scope of the With function, you can use With rather than Module, and gain a bit of speed.
You can perhaps try something like this:
Clear[f, primesList]
f = With[{p = Prime[#]},Piecewise[{{p, PrimeQ[p + 2]}}, {}] ] &;
primesList[k_] := Union#Flatten#(f /# Range[k]);
If you want both the prime p and the prime p+2, then the solution is
Clear[f, primesList]
f = With[{p = Prime[#]},Piecewise[{{p, PrimeQ[p + 2]}}, {}] ] &;
primesList[k_] :=
Module[{primes = f /# Range[k]},
Union#Flatten#{primes, primes + 2}];
Well, someone has to allocate memory somewhere for the full table size, since it is not known before hand what the final size will be.
In the good old days before functional programming :), this sort of thing was solved by allocating the maximum array size, and then using a separate index to insert to it so no holes are made. Like this
x=Table[0,{100}]; (*allocate maximum possible*)
j=0;
Table[ If[PrimeQ[k+2], x[[++j]]=k],{k,100}];
x[[1;;j]] (*the result is here *)
{1,3,5,9,11,15,17,21,27,29,35,39,41,45,51,57,59,65,69,71,77,81,87,95,99}
Here's another couple of alternatives using NextPrime:
pairs1[pmax_] := Select[Range[pmax], PrimeQ[#] && NextPrime[#] == 2 + # &]
pairs2[pnum_] := Module[{p}, NestList[(p = NextPrime[#];
While[p + 2 != (p = NextPrime[p])];
p - 2) &, 3, pnum]]
and a modification of your Reap/Sow solution that lets you specify the maximum prime:
pairs3[pmax_] := Module[{k,p},
Reap[For[k = 1, (p = Prime[k]) <= pmax, k++,
If[PrimeQ[p + 2], Sow[p]]]][[-1, 1]]]
The above are in order of increasing speed.
In[4]:= pairs2[10000]//Last//Timing
Out[4]= {3.48,1261079}
In[5]:= pairs1[1261079]//Last//Timing
Out[5]= {6.84,1261079}
In[6]:= pairs3[1261079]//Last//Timing
Out[7]= {0.58,1261079}

Maximize function in Mathematica which counts elements

I reduced a debugging problem in Mathematica 8 to something similar to the following code:
f = Function[x,
list = {1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5};
Count[list, x]
];
f[4]
Maximize{f[x], x, Integers]
Output:
4
{0, {x->0}}
So, while the maximum o function f is obtained when x equals 4 (as confirmed in the first output line), why does Maximize return x->0 (output line 2)?
The reason for this behavior can be easily found using Trace. What happens is that your function is evaluated inside Maximize with still symbolic x, and since your list does not contain symbol x, results in zero. Effectively, you call Maximize[0,x,Integers], hence the result. One thing you can do is to protect the function from immediate evaluation by using pattern-defined function with a restrictive pattern, like this for example:
Clear[ff];
ff[x_?IntegerQ] :=
With[{list = {1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5}}, Count[list, x]]
It appears that Maximize can not easily deal with it however, but NMaximize can:
In[73]:= NMaximize[{ff[x], Element[x, Integers]}, x]
Out[73]= {4., {x -> 4}}
But, generally, either of the Maximize family functions seem not quite appropriate for the job. You may be better off by explicitly computing the maximum, for example like this:
In[78]:= list = {1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5};
Extract[#, Position[#, Max[#], 1, 1] &[#[[All, 2]]]] &[Tally[list]]
Out[79]= {{4, 4}}
HTH
Try this:
list = {1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5};
First#Sort[Tally[list], #1[[2]] > #2[[2]] &]
Output:
{4, 4}

How to get the oddly indexed elements in a list in mathematica

How to get the oddly indexed elements in a list? I am thinking of Select, but did not find anything returning an element's position, especially considering there are repetitive elements in the list.
Also in general, how to select those elements whose indices satisfy some certain conditions?
Here's a few more in addition to #belisarius's answer, which don't require computing Length[lis]:
Take[lis, {1, -1, 2}]
lis[[1 ;; -1 ;; 2]]
You can often use -1 to represent the "last" position.
There are a lot of ways, here are some of them:
In[2]:= a = Range[10];le = Length#a;
In[3]:= Table[a[[i]], {i, 1, le, 2}]
In[5]:= Pick[a, Table[Mod[i, 2], {i, 1, le}], 1]
In[6]:= a[[1 ;; le ;; 2]]
In general, with Pick[] (as an example) you can model any conceivable index mask.
For some reason the terse form of Span has been omitted from the answers.
Range[20][[;;;;2]]
{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
Quoting the documentation:
;;;;k
from the beginning to the end in steps of k.

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