Re; newbie list question
- To: mathgroup at smc.vnet.net
- Subject: [mg115026] Re; newbie list question
- From: Richard Hofler <rhofler at bus.ucf.edu>
- Date: Tue, 28 Dec 2010 06:53:21 -0500 (EST)
Gareth,
I'm certain you'll get simpler solutions. However, here's one. . .
In[27]:== listA=={4,5,8,2,6,4};
listB=={8,4,2};
In[29]:== Flatten[Table[First[Position[{listA},_?(#====listB[[a]]&)]],{a,3}]/.{1,x_}->{x}]
Out[29]== {3,1,4}
This shows how most parts of the final function work.
Position[{listA},_?(#====listB[[2]]&)] (* example for only the 2nd element in ListB *)
First[Position[{listA},_?(#====listB[[2]]&)]] (* example for the first location of only the 2nd element in ListB *)
Table[First[Position[{listA},_?(#====listB[[a]]&)]],{a,3}] (* all of the elements in ListB *)
Table[First[Position[{listA},_?(#====listB[[a]]&)]],{a,3}]/.{1,x_}->{x}
Flatten[Table[First[Position[{listA},_?(#====listB[[a]]&)]],{a,3}]/.{1,x_}->{x}]
Out[37]== {{1,1},{1,6}}
Out[38]== {1,1}
Out[39]== {{1,3},{1,1},{1,4}}
Out[40]== {{3},{1},{4}}
Out[41]== {3,1,4}
HTH.
Richard Hofler
________________________________________
From: Gareth Edwards [gareth.edwards at cubicmotion.com]
Sent: Sunday, December 26, 2010 4:02 AM
To: mathgroup at smc.vnet.net
Subject: [mg115026] [mg114986] newbie list question
Hi,
Liking Mathematica a lot, but struggling with the early part of the learnin=
g curve, i.e. don't know what I don't know...
What would be the neatest syntax for finding the first location of elements=
from one list in another? For example:
listA == { 4,5,8,2,6,4 }
listB == { 8,4,2 }
I would like a function to return {3,1,4} in this case (the first location =
in A of each element in B)
Many thanks!