## string manipulation – ToExpression applied to an element of an array fails

When I apply `ToExpression` to a symbol, `a`, everthing works fine in that the string value of the symbol is converted to an Integer.

``````a = "1";

(* String *)

ToExpression(a)
(* 1 *)

(* Integer *)
``````

However when I apply it to an element of an array it fails

``````Clear(a)

a={{"name1", "‏‎11", "15", "AM"}, {"name2", "‏‎1", "31", "PM"}};

a((2,2))
(* 1 *)

(* String *)

ToExpression(a((2, 2)))
(* 1 *)

(* Symbol *)
``````

I discovered this when attempting something like:

``````ToExpression(a((2, 2))) + 12
(* 12 + 1 *)
``````

I got the indicated answer rather than the Integer, 13.

## list manipulation – Export data to file

I have a function f(x,y,z,i) that returns the solutions {x1,y1,z1} for each step “i”.
I would like to use a repeating command (such as Do(…)) to get the outputs of this function f saved in a file. More specifically, I thought using something like Do(f(x,y,z,i),{i,1,10,1}) combined with Save, but I do not know how to combine these two commands with the interation of “i”

## list manipulation – How to construct a tree from a preorder traversal

The list

``````t1 = {1, 2, 3, 3, 4, 4, 3, 3, 4, 3, 3, 2, 3, 4, 2, 3, 2};
``````

might be derived from a pre-order traversal of

``````tree1 = Tree[1, {Tree[2, {3, Tree[3, {4, 4}], 3, Tree[3, {4}], 3, 3}],
Tree[2, {Tree[3, {4}]}], Tree[2, {3}], 2}]
``````

But how to take the original list, `t1` here, and build the tree `tree1` ?

What have I tried so far ? Nothing to any avail! Well, that’s not quite true, I have made some progress with a very procedural code to walk along the list and try to figure out the structure of the tree as each new element is read, but I have a dispiriting feeling that I have missed an obvious usage of some of the smarter functional programming functionality.

## list manipulation – How to evaluate a function on all solutions given in output of a Reduce function

Say I have a Mathematica programme/function $$f(z_,w_{} ) = mathrm{Reduce}(cdots)$$ which outputs a list of solutions, which looks something like $$a = 1 && ((0leq b leq 1000 && -5 leq c leq 65))$$
$$a = 2 && ((32leq b leq 860 && -4 leq c leq 36))$$
and so on for all cases of $$a$$ (here $$1 leq a leq 10$$). Here $$a,b,c$$ are all integers.

I have another function $$g$$ written in Mathematica taking values in four integers that I would like to evaluate on each solution in the list above. Is it possible to write code which does this in Mathematica? I have searched online but haven’t had much luck. Thanks!

## list manipulation – Why are the inputs not being applied to the function in this MWE?

Consider the following MWE

``````Clear("Global`*")

randFunc(listFunc_) := RandomChoice(listFunc)@x;
mwe(list_) := Module({testfun1, testfun2},
testfun1(x_) = RandomChoice(list)@x;
testfun2(x_) = randFunc(list);

N@{testfun1(2), testfun2(2)}
);

mwe({Sqrt, Log})
``````

Which gives output that looks like `{0.693147, Log(x)}`

The numbers/function changes because of `RandomChoice`, but why does testfun1(2) evaluate but not testfun2(2)?

Note: The following code (where the list is explicitly input in `testfun1` and `testfun2` does work

Clear(“Global`*”)

``````randFunc(listFunc_) := RandomChoice(listFunc)@x;
testfun1(x_, list_) = RandomChoice(list)@x;
testfun2(x_, list_) = randFunc(list);
N@{testfun1(2, {Sqrt, Log}), testfun2(2, {Sqrt, Log})}
``````

Which gives output like `{1.41421, 0.693147}` (again, `RandomChoice` can change values, but they are both evaluating)

Note 2: I can get the expected behavior in the first example by changing `testfun2(x_) = randFunc(list) ;`to `testfun2(y_) = randFunc(list) /. x -> y;`, but I don’t understand why I would need to do this.

Looking at the trace, Maybe `randFunc(list)` is not evaluating when `testfun2(x_)` is being defined in the `Module`, but I don’t understand why this would be?

• (maybe the `HoldAll` attribute of `Module`?)

## list manipulation – Indexed[] returning Table::nliter error

I have been trying to create a new table from a list of values. I used

``````testtab = {{1, 1}, {2, 1}, {3, 4}, {5, 10}};
difftesttab =
Table(Indexed(testtab, {part, 1}), (
Indexed(testtab, {part + 1, 2}) - Indexed(testtab, {part, 2}))/(
0.5 (Indexed(testtab, {part + 1, 2}) +
Indexed(testtab, {part, 2}))), {part, 1, Length(testtab) - 1})
``````

which gives me a Table::nliter error. I also tried using testtab((part,1)) instead of Indexed, but that gives me a Part::pkspec1 error. Do you have any suggestions on how to approach this?

TIA!

## list manipulation – Diagonal matrix command in Mathematica

``````AA=rand(3,3)
tmp = spdiags(AA,0)
tmp(2)=1
backmat=spdiags(tmp,0,AA);
full(backmat)
``````

gives

``````AA =
0.6948    0.0344    0.7655
0.3171    0.4387    0.7952
0.9502    0.3816    0.1869

tmp =
0.6948
0.4387
0.1869

tmp =
0.6948
1.0000
0.1869

ans =
0.6948    0.0344    0.7655
0.3171    1.0000    0.7952
0.9502    0.3816    0.1869
``````

In Mathematica

``````(AA = {{0.6948, 0.0344, 0.7655}, {0.3171, 0.4387, 0.7952}, {0.9502,
0.3816, 0.1869}}) // MatrixForm
tmp = Diagonal(AA, 0);
tmp((2)) = 1;
backmat =
SparseArray(Band({1, 1}) -> tmp) +
SparseArray@UpperTriangularize(AA, 1) +
SparseArray@LowerTriangularize(AA, -1)
MatrixForm(backmat)
``````

Gives

I do not think Mathematica has command to insert diagonal into sparse matrix directly like Matlab’s spdiags but it is possible to do it as above indirectly.

## algebraic manipulation – Rearranging inequality

I’m trying to rearrange this inequality $$a + b a – 1 > a^2 – c a – b c$$, to appear as $$-a^2 + (b+c+1)a>1-bc$$ or $$-a^2 + (b+c+1)a-1+bc>0$$. I’ve tried to achieve that with `Collect`, but couldn’t make it work on both sides of the inequality.

Any ideas on how to achieve that?

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There’s a list such as `{0.1,2,0.15,5}`. I want to know how to convert the decimal to integer( but begin with zero `0`), and don’t touch the integer. In the second step, joint the data to a file name “a01b2c015d5”. Note that the data in the printed file name are separated with a,b,c and d, because I have tons of data to mark with a, b, c and d. How to do it? Thank you!