Re: Using Select with arrays? (Relative newbie)

*To*: mathgroup at smc.vnet.net*Subject*: [mg54258] Re: Using Select with arrays? (Relative newbie)*From*: János <janos.lobb at yale.edu>*Date*: Mon, 14 Feb 2005 21:50:53 -0500 (EST)*References*: <200502140317.WAA14058@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Well, I did not read to the end at first. If you want to use Select to select the element which has maximum in its #2 location from all those who have 1 in their #1 location, then you can do something like: In[23]:= Select[Select[data, #1[[1]] == 1 & ], #1[[2]] == Max[ Select[data, #1[[1]] == 1 & ][[All, 2]]] & ] Out[23]= {{1, 4, string4, c, d}} János On Feb 13, 2005, at 10:17 PM, Hugo Mallinson wrote: > The subject might not be entirely correct, but here is my problem: > I have a list of 5-variable data points like > data = { > {1, 1, string1, c, d} > {1, 2, string2, c, d} > {1, 3, string3, c, d} > {1, 4, string4, c, d} > {2, 1, string1, c, d} > {2, 2, string2, c, d} > {3, 1, string3, c, d} > {4, 1, string2, c, d} > {4, 2, string4, c, d} > } > and I want to extract just the points that have 1 (or 2, etc) as their > first value. I think I should do something like > Select[data, {1, _Integer, _String, _Integer, _Integer}] > but that doesn't work. > Having done that I need to find the maximum value of #2 for each > string, which I presumably do by the same method as above to extract > all string1 (or ...2) and then use Map[] and Max[]. I would do this all > with For loops (revealing my lack of Mathematica chops :-) ) but I'd > really like to learn how to do this sort of extraction in a > Mathematica-y way. Any help would be greatly appreciated! > Hugo ---------------------------------------------- Trying to argue with a politician is like lifting up the head of a corpse. (S. Lem: His Master Voice)

**Follow-Ups**:**Re: Re: Using Select with arrays? (Relative newbie)***From:*DrBob <drbob@bigfoot.com>

**References**:**Using Select with arrays? (Relative newbie)***From:*Hugo Mallinson <hfm21@cam.ac.uk>