Re: Question about thread

*Subject*: [mg2643] Re: [mg2581] Question about thread*From*: hay at haystack.demon.co.uk (Allan Hayes)*Date*: Thu, 30 Nov 1995 21:02:50 -0500*Approved*: usenet@wri.com*Distribution*: local*Newsgroups*: wri.mathgroup*Organization*: Wolfram Research, Inc.

sherod at boussinesq.Colorado.EDU (Scott Herod) in [mg2581] Question about thread writes "I have a question about the way Thread behaves. From the description in the _Mathematica_ book, I would expect it to return lists. For example " (attached) Scott, 1.Thread is not restricted to Lists Thread[f[h[1,2],h[3,4]], h ] (*use 2nd place*) h[f[1, 3], f[2, 4]] 2.We have some control where the threading takes place Thread[f[x,g[y,z],h[a1,b1],h[a2,b2]],h,3] (*use 2nd & 3rd places*) h[f[x, g[y, z], a1, h[a2, b2]], f[x, g[y, z], b1, h[a2, b2]]] 3.The behaviour of your example: blat[p_, q_] := Module[{}, Plus @@ (Variables[p]^q)] Thread[blat[{a+c,b+d},x]] x x x x a + b + c + d is due to blat being evaluated before Thread acts. Here are some ways that this can be prevented. Thread[Unevaluated[blat[{a+c,b+d},x]]] x x x x {a + c , b + d } ReleaseHold[Thread[Hold[blat][{a+c,b+d},x]]] x x x x {a + c , b + d } Block[{blat},Thread[blat[{a+c,b+d},x]]] x x x x {a + c , b + d } Where it is suitable the first way is usually quickest. The last one depends on Literal[blat] being a symbol] Allan Hayes hay at haystack.demon.co.uk ******** Begin forwarded message: >From: sherod at boussinesq.Colorado.EDU (Scott Herod) >Subject: [mg2581] Question about thread I have a question about the way Thread behaves. From the description in the _Mathematica_ book, I would expect it to return lists. For example Mathematica 2.2 for Solaris Copyright 1988-93 Wolfram Research, Inc. License valid through 28 Nov 1995. -- Open Look graphics initialized -- In[1]:= Thread[f[{a,b},x]] Out[1]= {f[a, x], f[b, x]} But that doesn't always seem to be the case. In[2]:= blat[p_, q_] := Module[{}, Plus @@ (Variables[p]^q)] In[3]:= blat[a+b,x] x x Out[3]= a + b In[4]:= Thread[blat[{a+c,b+d},x]] x x x x Out[4]= a + b + c + d Why isn't Out[4] the same as In[5]:= Outer[blat, {a+c, b+d} , {x}] // Flatten x x x x Out[5]= {a + c , b + d } Scott Herod Applied Mathematics University of Colorado, Boulder sherod at colorado.edu