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Re: confusion about Thread[]


Hi Torsten,

see below

Torsten Coym wrote:
> Hi group,
> 
> given three lists
> 
> wLst = {w1, w2, w3}
> zLst = {z1, z2, z3}
> cLst = {c1, c2, c3}
> 
> with each element representing a boolean value I want to calculate a 
> list of the same function applied to the corresponding elements of wLst, 
> zLst, cLst respectively. While the expression
> 
> 
> Thread[Thread[wLst && zLst] || Thread[wLst && cLst] ||
>     Thread[zLst && cLst]]
> 
> 
> {(w1 && z1) || (w1 && c1) || (z1 && c1), (w2 && z2) || (w2 && c2) ||
>     (z2 && c2), (w3 && z3) || (w3 && c3) || (z3 && c3)}
> 
> does exactly what I want, I wonder why the following approach using a 
> pure function with three input arguments and a single call of Thread[] 
> does not give the desired result:
> 
> 
> Thread[((#1 && #2) || (#1 && #3) || (#2 && #3) & )[wLst, zLst, cLst]]
> 
> ({w1, w2, w3} && {z1, z2, z3}) || ({w1, w2, w3} && {c1, c2, c3}) ||
>    ({z1, z2, z3} && {c1, c2, c3})
You simply forgot the inner Thread:

Thread[(Thread[(#1 && #2)] || Thread[(#1 && #3)] || Thread[(#2 && #3)] 
&)[wLst, zLst, cLst]]

SIncerely, Daniel

> 
> although
> 
> ((#1 && #2) || (#1 && #3) || (#2 && #3) & )[w1, z1, c1]
> 
> (w1 && z1) || (w1 && c1) || (z1 && c1)
> 
> implements the desired logical expression and
> 
> 
> Thread[f[wLst, zLst, cLst]]
> 
> {f[w1, z1, c1], f[w2, z2, c2], f[w3, z3, c3]}
> 
> works the way I expect it.
> 
> 
> It seems I am a bit disconnected here ...
> 
> Torsten
> 


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