Zsbán, I learned a lot along the way. First "Not is Nice": D=:'1234:ABC' f=: 13 :'y-.'':''' f D 1234ABC f ':' -.~ ] Second "Easy alphanumeric characters"
]L=:(2|'/9@Z`z'I.a.)#a. 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz However, all the side trips along the way were fun. Linda On 3/8/13, Linda Alvord <[email protected]> wrote: > ]A=:1 2 3 { 8 32 $ a. NB. Example from definition of a. > ]AL=:_5}."1 1}."1 A > ]L=:15}.,AL > f=: 13 :'((i.#y)~:y i. '':'')#y' NB. Remove the : from list > f > ] #~ ([: i. #) ~: ':' i.~ ] > f L > 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz Why don't you use this instead? f=: -.&':' In any case, this is a nice method, and with some liberal changes, can be shortened to this. 42}.,4{._32(26&{.)\a.-.':' (Incidentally, I'm collecting these expressions on "http://www.jsoftware.com/jwiki/BJonas";. > Here's where I got lost in your solution: > > a.#~2|a.i.&0@/:@,"{'/9@Z`z' > > I can't seem to unlock what happens at the heart of it. > What does this do? > > ([: i.&0 /:) I admit that solution is a bit overcomplicated. It's not what I'd write in a real program. Let me explain how it works. This post will be a bit long, sorry, but this is the easiest way I can explain. I'll show you a much easier solution at the end though. I'll show on a shorter range of characters for readability. Suppose we start from this range of characters. ]alp =: (26{.65}.a.) ABCDEFGHIJKLMNOPQRSTUVWXYZ Take this key of six characters which we'll use as endpoints of three intervals to cut the above. ]key=: 'DHMQTV' DHMQTV Suppose we concatenate each letter with the key and sort the resulting seven character long string. alp /:~@,"0 1 key ADHMQTV BDHMQTV CDHMQTV DDHMQTV DEHMQTV DFHMQTV DGHMQTV DHHMQTV DHIMQTV DHJMQTV DHKMQTV DHLMQTV DHMMQTV DHMNQTV DHMOQTV DHMPQTV DHMQQTV DHMQRTV DHMQSTV DHMQTTV DHMQTUV DHMQTVV DHMQTVW DHMQTVX DHMQTVY DHMQTVZ As you can see, the single letter we concatenate goes to different positions in the seven letter string, and jumps to a new position whenever we reach a letter in the key. For example, ABCD are in the first position in the first four lines, then EFGH are in the second position, then IJKLM in the third. The boundaries are exactly DHM because those are the first three letters of the key. Now instead of sorting the seven letter strings, we grade them to find out where the letter we insert goes. alp /:@,"0 1 key 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 1 0 2 3 4 5 6 1 0 2 3 4 5 6 1 0 2 3 4 5 6 1 0 2 3 4 5 6 1 2 0 3 4 5 6 1 2 0 3 4 5 6 1 2 0 3 4 5 6 1 2 0 3 4 5 6 1 2 0 3 4 5 6 1 2 3 0 4 5 6 1 2 3 0 4 5 6 1 2 3 0 4 5 6 1 2 3 0 4 5 6 1 2 3 4 0 5 6 1 2 3 4 0 5 6 1 2 3 4 0 5 6 1 2 3 4 5 0 6 1 2 3 4 5 0 6 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 As you can see, the position of the zero in this string shows where the extra letter got inserted. So let's find the position of the zero, putting the corresponding inserted letter above for clarity. |:alp;"0 i.&0"1 alp /:@,"0 1 key +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ |A|B|C|D|E|F|G|H|I|J|K|L|M|N|O|P|Q|R|S|T|U|V|W|X|Y|Z| +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ |0|0|0|0|1|1|1|1|2|2|2|2|2|3|3|3|3|4|4|4|5|5|6|6|6|6| +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ So now the characters up to D get converted to a zero, the characters up to H get converted to a one, etc. The key string controls where exactly the numeric result changes. So if we take the letters that get converted to an odd number, we get exactly the letters corresponding to the three intervals determined by the key string: EFGH for the first interval (between D and H), then NOP for the second interval (from M to P), etc. alp#~ 2| i.&0"1 alp /:@,"0 1 key EFGHNOPQUV In the actual solution to get the alphanumeric characters, I use '/9@Z`z' as the key because '/9' gives the interval for numeric characters, '@Z' the one for uppercase letters, etc; and I use a. as the starting list of letters. Now it turns out the whole sort thing is a bit superfluous, because the search primitive gives a much simpler and shorter solution. alp#~ 2| key I. alp EFGHNOPQUV a.#~2|'/9@Z`z'I.a. 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz I wonder why I didn't find that earlier. -- Ambrus ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
