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I have a table form as:

table = {{a, 6}, {b, ff}, {c, 2}}

I want to associate the first element of each pair from the table to the second element of the pair as in:

a=6, b=ff, c=2.

I want the parameters a,b,c to be recognized throughout the program with assigned values. Doing as below, everything is perfect:

a := 6; a^2 + 5 Clear[a] Out[2] 41

But in the example below, I failed:

Clear[table, a, b, c] table = {{a, 6}, {b, ff}, {c, 2}} z = table[[1, 1]] z := table[[1, 2]] a^2 + 5 Out[4] {{a, 6}, {b, ff}, {c, 2}} Out[5] a Out[6] 5 + a^2

Why this fail?

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    $\begingroup$ I don't see why it should work. One way to assign the second element to the first element is Set @@@ {{a, 6}, {b, ff}, {c, 2}}. Now a returns 6, b returns ff and so on. Maybe if you explain why you think it should work, someone could explain why it doesn't. $\endgroup$ Commented Jan 9, 2014 at 15:11

2 Answers 2

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This may be considered a duplicate of Assign the results from a Solve to variable(s) which despite different formulation shares the same shortest answer:

Set @@@ {{a, 6}, {b, ff}, {c, 2}}

This may also be related to Reassign values to symbols if you expect to be able to make different assignments to the same Symbols in the same way. I mean that if you attempt a second series of assignments in the same manner it will not work:

Set @@@ {{a, 5}, {b, Pi}, {c, 7/3}}

Set::setraw: Cannot assign to raw object 6. >>

Set::setraw: Cannot assign to raw object 2. >>

An alternative form that will work requires keeping the assignment pairs in Hold:

List @@ Set @@@ Hold[{a, 5}, {b, Pi}, {c, 7/3}]; {a, b, c} 
{5, π, 7/3}

Also related:

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  • $\begingroup$ Thank you very much for your answer. I'll keep in mind the problems that arise from reassign symbols. $\endgroup$ Commented Jan 10, 2014 at 9:01
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SetDelayed has attribute HoldAll, so your z := table[[1, 2]] assigns the righthand-side to z instead of a. If you really insist on doing your assignment in this programmatic way, use Evaluate and Set:

Clear[table, a, b, c] table = {{a, 6}, {b, ff}, {c, 2}} z = table[[1, 1]] Evaluate[z] = table[[1, 2]]

Or SetDelayed, but then expect a to be recalculated every time in table which leads to a circular definition of a, causing infinite recursion. Hence the Quiet:

Clear[table, a, b, c] table = {{a, 6}, {b, ff}, {c, 2}} z = table[[1, 1]] Evaluate[z] := Quiet@table[[1, 2]]

A more elegant way (as pointed out by Anon) is to use Set and Apply:

Clear[table, a, b, c]; table = {{a, 6}, {b, ff}, {c, 2}}; Set @@@ table
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  • $\begingroup$ Thank you very much for your proposed solution which helped me completely. Very elegant indeed, with Set and Apply $\endgroup$ Commented Jan 10, 2014 at 8:55

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