I basically want the same thing as this OP:
Is there a J idiom for adding to a list until a certain condition is met?
But I cant get the answers to work with OP's function or my own.
I will rephrase the question and write about the answers at the bottom.
I am trying to create a function that will return a list of fibonacci numbers less than 2.000.000. (without writing "while" inside the function).
Here is what i have tried:
First, i picked a way to culculate fibonacci numbers from this site:
https://code.jsoftware.com/wiki/Essays/Fibonacci_Sequence
fib =: (i. +/ .! i.#-)"0
echo fib i.10
0 1 1 2 3 5 8 13 21 34
Then I made an arbitrary list I knew was larger than what I needed. :
fiblist =: (fib i.40) NB. THIS IS A BAD SOLUTION!
Finally, I removed the numbers that were greater than what I needed:
result =: (fiblist < 2e6) # fiblist
echo result
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1.34627e6
This gets the right result, but is there a way to avoid using some arbitrary number like
40 in "fib i.40" ?
I would like to write a function, such that "func 2e6" returns the list of fibonacci numbers below 2.000.000. (without writing "while" inside the function).
echo func 2e6
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1.34627e6
here are the answers from the other question:
first answer:
2 *^:(100&>#:])^:_"0 (1 3 5 7 9 11)
128 192 160 112 144 176
second answer:
+:^:(100&>)^:(<_) ] 3
3 6 12 24 48 96 192
As I understand it, I just need to replace the functions used in the answers, but i dont see how
that can work. For example, if I try:
echo (, [: +/ _2&{.)^:(100&>#:])^:_ i.2
I get an error.
I approached it this way. First I want to have a way of generating the nth Fibonacci number, and I used f0b from your link to the Jsoftware Essays.
f0b=: (-&2 +&$: -&1) ^: (1&<) M.
Once I had that I just want to put it into a verb that will check to see if the result of f0b is less than a certain amount (I used 1000) and if it was then I incremented the input and went through the process again. This is the ($:#:>:) part. $: is Self-Reference. The right 0 argument is the starting point for generating the sequence.
($:#:>: ^: (1000 > f0b)) 0
17
This tells me that the 17th Fibonacci number is the largest one less than my limit. I use that information to generate the Fibonacci numbers by applying f0b to each item in i. ($:#:>: ^: (1000 > f0b)) 0 by using rank 0 (fob"0)
f0b"0 i. ($:#:>: ^: (1000 > f0b)) 0
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987
In your case you wanted the ones under 2000000
f0b"0 i. ($:#:>: ^: (2000000 > f0b)) 0
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269
... and then I realized that you wanted a verb to be able to answer your original question. I went with dyadic where the left argument is the limit and the right argument generates the sequence. Same idea but I was able to make use of some hooks when I went to the tacit form. (> f0b) checks if the result of f0b is under the limit and ($: >:) increments the right argument while allowing the left argument to remain for $:
2000000 (($: >:) ^: (> f0b)) 0
32
fnum=: (($: >:) ^: (> f0b))
2000000 fnum 0
32
f0b"0 i. 2000000 fnum 0
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269
I have little doubt that others will come up with better solutions, but this is what I cobbled together tonight.
Related
There have been numerous questions about inconsistent results from the YouTube Data API: 1, 2, 3, 4, 5, 6. Most of them have accepted answers that seem to indicate there was a problem with the API request that was fixed by the instructions in the answers. But none of those situations apply to the API request discussed here.
There have also been two questions about duplicates in the API results: 1, 2. Both of them have the same answer, which says to use the next-page token. But both questions say the token was used, so that answer is not helpful.
Yesterday, I submitted a series of API requests to get the list of most-viewed videos about 3D printing. The first request in the series was:
https://www.googleapis.com/youtube/v3/search?q=3D print&type=video&maxResults=50&part=id,snippet&order=viewCount&key=<my key>
I ran that in a VBA sub, which took the next-page token from each result and resubmitted the URL with &pageToken=<nextPageToken> inserted.
The result was a list of 649 unique video IDs. So far so good.
After making some changes in the VBA code and seeing some duplicates in the result set, I went back today and ran the original VBA sub again. The result was again a list of 649 video IDs, but this time the list included duplicates and it also included IDs that were not in yesterday's list and was missing IDs that were there yesterday. Here is a comparison from the first two pages and the last two pages of the two result sets:
Page
# on page
# overall
Run 1
Run 2
Same as
Seq
Dup
1
1
1
f2mdMcf-fJs
f2mdMcf-fJs
1
1
2
2
WSauz5KVKTU
WSauz5KVKTU
2
Seq
1
3
3
zsSCUWs7k9Q
XYIUM5TkhMo
None
1
4
4
B5Q1J5c8oNc
zsSCUWs7k9Q
3
Seq
1
5
5
cUxIb3Pt-hQ
B5Q1J5c8oNc
4
Seq
1
6
6
4yyOOn7pWnA
LDjE28szwr8
None
1
7
7
3N46jQ0Xi3c
cUxIb3Pt-hQ
5
Seq
1
8
8
08dBVz8_VzU
4yyOOn7pWnA
6
Seq
...
1
13
13
oeKIe1ik2O8
e1rQ8YwNSDs
11
Seq
1
14
14
FrG_eSECfps
RVB2JreIcoc
12
Seq
1
15
15
pPQCwz2q96o
oeKIe1ik2O8
13
Seq
1
16
16
uo3KuoEiu3I
pPQCwz2q96o
15
NOT
1
17
17
0U6aIwd5h9s
uo3KuoEiu3I
16
Seq
...
1
47
47
ShGsW68zbIo
iu9rhqsvrPs
46
Seq
1
48
48
0q0xS7W78KQ
ShGsW68zbIo
47
Seq
1
49
49
UheJQsXOAnY
0q0xS7W78KQ
48
Seq
Dup
1
50
50
H8AcqOh0wis
H8AcqOh0wis
50
NOT
Dup
2
1
51
EWq3-2VuqbQ
0q0xS7W78KQ
48
NOT
Dup
2
2
52
scuTZza4f_o
H8AcqOh0wis
50
NOT
Dup
2
3
53
bJWJW-mz4_U
UheJQsXOAnY
49
NOT
2
4
54
Ii4VYsh9OlM
EWq3-2VuqbQ
51
NOT
2
5
55
r2-OGUu57pU
scuTZza4f_o
52
Seq
2
6
56
8KTnu18Mi9Q
bJWJW-mz4_U
53
Seq
2
7
57
DconsfGsXyA
Ii4VYsh9OlM
54
Seq
2
8
58
UttEvLJP3l8
8KTnu18Mi9Q
56
NOT
2
9
59
GJOOLH9ZP2I
DconsfGsXyA
57
Seq
2
10
60
ewgmg9Q5Ab8
UttEvLJP3l8
58
Seq
...
13
35
635
qHpR_p8lA4I
FFVOzo7tSV8
639
Seq
13
36
636
DplwDDZNTRc
76IBjdM9s6g
640
Seq
13
37
637
3AObqGsimr8
qEh0uZuu7_U
None
13
38
638
88keQ4PWH18
RhfGJduOlrw
641
Seq
13
39
639
FFVOzo7tSV8
QxzH9QkirCU
643
NOT
13
40
640
76IBjdM9s6g
Qsgz4GbL8O4
None
13
41
641
RhfGJduOlrw
BSgg7mEzfqY
644
Seq
13
42
642
lVEqwV0Nlzg
VcmjbJ2q8-w
645
Seq
13
43
643
QxzH9QkirCU
gOU0BCL-TXs
None
13
44
644
BSgg7mEzfqY
IoOXQUcW24s
646
Seq
13
45
645
VcmjbJ2q8-w
o4_2_a6LzFU
647
Seq
Dup
14
1
646
IoOXQUcW24s
o4_2_a6LzFU
647
NOT
Dup
14
2
647
o4_2_a6LzFU
ijVPcGaqVjc
648
Seq
14
3
648
ijVPcGaqVjc
nk3FlgEuG-s
649
Seq
14
4
649
nk3FlgEuG-s
27ZLFn8Dejg
None
The last three columns have the following meanings:
Same as: If an ID from Run 2 is the same as an ID from Run 1, then this column has the # overall for Run 1.
Seq: Indicates whether the number in column "Same as" is one more than the previous number in that column.
Dup: Indicates whether an ID from Run 2 occurred more than once in that run.
Problems:
The videos XYIUM5TkhMo, LDjE28szwr8, qEh0uZuu7_U, Qsgz4GbL8O4, gOU0BCL-TXs, and 27ZLFn8Dejg were returned as #3, 6, 637, 640, 643, and 649 in Run 2, but were not returned at all in Run 1.
The videos FrG_eSECfps, r2-OGUu57pU, lVEqwV0Nlzg were returned as #14, 55, 642, in Run 1, but were not in Run 2.
The videos 0q0xS7W78KQ, H8AcqOh0wis, and o4_2_a6LzFU were returned as #49, 50, and 645 in Run 2, but then each appears a second time in that run (as well as appearing in Run 1 as #48, 50, and 647).
These results are troubling. They mean that no single search will return a reliable list of videos for a given value of q.
I mentioned at the beginning that previous questions about inconsistent results from the YouTube Data API had answers that seemed to resolve those inconsistencies. Is there a way to do that for this search? Is there something wrong with the way I'm composing the search that is causing the problem?
If there isn't a way to fix the search, then I suppose the only way to get a list of videos on the topic with high confidence of it being complete is to run the search multiple times and merge the results until no new IDs appear that were not in a previous result set. But even then, one would not know if there are other videos lurking undetected.
I'm looking for the exact specifications of this file format. Anyone got a link? Or want to comment?
I have spent the better part of the day searching, yet I keep getting directed back to the GIMP online user-manual. It says "look at a .gpl file and you will see it is easy" to build manually with a text editor. I don't actually have GIMP, but I see examples online. Yep, easy. • EXCEPT:
• What meaning do the color names ultimately have? Are they purely semantic, or does a program rely on them? If the latter, then what if there are two (2) or more colors with the same name?
• What does the "Columns" line do?
I've seen examples that have no "Columns" line.
I've seen examples that have values of 0, 4, and 16; yet this does not in any way that I can see correspond to the color data. I see 3 columns of decimal-sRGB values, and an optional 4th column with the color-name; seems I remember the example with "Columns 4" had no color names, only the 3 RGB columns.
• Do columns of RGB values need to "line up"? Or will the following example from my output algorithm work? (from the Crayola palette):
159 129 112 Beaver
253 124 110 Bittersweet
0 0 0 Black
172 229 238 Blizzard Blue
31 117 254 Blue
162 162 208 Blue Bell
102 153 204 Blue Gray
13 152 186 Blue Green
• Does this format accept sRGBA colors? And if so, how is the "A" value defined (0-1, 0%-100%, 0-127, 0-255, etc.?) (seems I remember when creating .png files with PHP, the "A" value was 7-bit)?
• How exactly do you add comments / metadata?
Today I see an example that says lines that begin with # are comments, or anything after a # on a line is a comment. Yesterday I thought (maybe I'm confused) I saw an example that said that comment lines begin with ;
• Is any other data-format supported?
Originally I thought the text-line just before the color-data that I see in every example indicated the format: "#" signifying decimal-sRGB; until today when I see that is just a blank-line comment.
• What line ending character(s) can / must I use?
\n
\r
• What character-encodings can I use? ASCII only? ¿UTF-8 ☺ with extended ♪♫ charset (¡hopefully!)?
• Anything I'm missing? Any other options available?
Here is an example from http://gimpchat.com/viewtopic.php?f=8&t=3375#
GIMP Palette
Name: bugslife_final.png-10
Columns: 16
#
191 180 180 Index 0
163 158 157 Index 1
145 136 132 Index 2
130 125 112 Index 3
… … …
56 50 49 Index 29
41 38 38 Index 30
23 23 23 Index 31
242 245 213 Index 32
227 232 181 Index 33
210 217 147 Index 34
195 204 118 Index 35
… … …
0 0 0 Index 251
0 0 0 Index 252
0 0 0 Index 253
0 0 0 Index 254
0 0 0 Index 255
Aloha!
Looking at the source code:
Columns is just an indication for display in the palette editor
Comments must start with a #. In non-empty lines that don't, the first three tokens are parsed as numbers
There is no alpha support
I have following code now, which stores the indices with the maximum score for each question in pred, and convert it to string.
I want to do the same for n-best indices for each question, not just single index with the maximum score, and convert them to string. I also want to display the score for each index (or each converted string).
So scores will have to be sorted, and pred will have to be multiple rows/columns instead of 1 x nqs. And corresponding score value for each entry in pred must be retrievable.
I am clueless as to lua/torch syntax, and any help would be greatly appreciated.
nqs=dataset['question']:size(1);
scores=torch.Tensor(nqs,noutput);
qids=torch.LongTensor(nqs);
for i=1,nqs,batch_size do
xlua.progress(i, nqs)
r=math.min(i+batch_size-1,nqs);
scores[{{i,r},{}}],qids[{{i,r}}]=forward(i,r);
end
tmp,pred=torch.max(scores,2);
answer=json_file['ix_to_ans'][tostring(pred[{i,1}])]
print(answer)
Here is my attempt, I demonstrate its behavior using a simple random scores tensor:
> scores=torch.floor(torch.rand(4,10)*100)
> =scores
9 1 90 12 62 1 62 86 46 27
7 4 7 4 71 99 33 48 98 63
82 5 73 84 61 92 81 99 65 9
33 93 64 77 36 68 89 44 19 25
[torch.DoubleTensor of size 4x10]
Now, since you want the N best indexes for each question (row), let's sort each row of the tensor:
> values,indexes=scores:sort(2)
Now, let's look at what the return tensors contain:
> =values
1 1 9 12 27 46 62 62 86 90
4 4 7 7 33 48 63 71 98 99
5 9 61 65 73 81 82 84 92 99
19 25 33 36 44 64 68 77 89 93
[torch.DoubleTensor of size 4x10]
> =indexes
2 6 1 4 10 9 5 7 8 3
2 4 1 3 7 8 10 5 9 6
2 10 5 9 3 7 1 4 6 8
9 10 1 5 8 3 6 4 7 2
[torch.LongTensor of size 4x10]
As you see, the i-th row of values is the sorted version (in increasing order) of the i-th row of scores, and each row in indexes gives you the corresponding indexes.
You can get the N best values/indexes for each question (i.e. row) with
> N_best_indexes=indexes[{{},{indexes:size(2)-N+1,indexes:size(2)}}]
> N_best_values=values[{{},{values:size(2)-N+1,values:size(2)}}]
Let's see their values for the given example, with N=3:
> return N_best_indexes
7 8 3
5 9 6
4 6 8
4 7 2
[torch.LongTensor of size 4x3]
> return N_best_values
62 86 90
71 98 99
84 92 99
77 89 93
[torch.DoubleTensor of size 4x3]
So, the k-th best value for question j is N_best_values[{{j},{values:size(2)-k+1}]], and its corresponding index in the scores matrix is given by this row, column values:
row=j
column=N_best_indexes[{{j},indexes:size(2)-k+1}}].
For example, the first best value (k=1) for the second question is 99, which lies at the 2nd row and 6th column in scores. And you can see that values[{{2},values:size(2)}}] is 99, and that indexes[{{2},{indexes:size(2)}}] gives you 6, which is the column index in the scores matrix.
Hope that I explained my solution well.
This program insists that 35 is a prime number even though, going through it step-by-step, the program should reach the point where it calculates 35%5 and then ignore the number (because the result is 0.) I haven't checked every single number but it seems to display only primes otherwise (except for numbers that are anologous to 35 like 135.)
print ('How many prime numbers do you require?')
primes = io.read("*n")
print ('Here you go:')
num,denom,num_primes=2,2,0
while num_primes<primes do
if denom<num then
if num%denom==0 then
num=num+1
else
denom=denom+1
end
else
print(num)
num=num+1
num_primes=num_primes+1
denom=2
end
end
Sample output:
How many prime numbers do you require?
50
Here you go:
2
3
5
7
11
13
17
19
23
27
29
31
35
37
41
43
47
53
59
61
67
71
73
79
83
87
89
95
97
101
103
107
109
113
119
123
127
131
135
137
139
143
147
149
151
157
163
167
173
179
You aren't resetting denom in the % case.
if num%denom==0 then
num=num+1
else
So when you fall-through this test you start testing the next number starting from the previous denominator instead of from 2 again.
Simple debugging print lines in the loop printing out denom and num would have shown this to you (as, in fact, that's exactly how I found it). You only need to three prime numbers output to see the issue.
Fixed it, set denom=2 after num=num+1
print ('How many prime numbers do you require?')
primes = io.read("*n")
print ('Here you go:')
num,denom,num_primes=2,2,0
while num_primes<primes do
if denom<num then
if num%denom==0 then
num=num+1
denom=2
else
denom=denom+1
end
else
print(num)
num=num+1
num_primes=num_primes+1
denom=2
end
end
HI complete newbie question here: I have a table consisting of two columns. First column belongs to "bins" that are coded by where a the fruit flies live. The second column is either 0 or 1, neutral vs really like sugar, respectively. I have two question?
1) if I suspect that there is a single variable, something about where they live that is determining whether how much they like sugar. Is there a way that I can have the computer to group into just 2 clusters? All the bins that like sugar vs neutral. That way we can do further experiment to determine what is it about the bins.
2) automatically determine how many clusters there might be that is driving this behavior? For example may be there is 4 variables (4 clusters) that can determine the outcome of sugar preference.
Apologies if this is trivial. The table is listed below. thanks!
Bin sugar
1 1
1 1
1 0
1 0
2 1
2 0
2 0
3 1
3 0
3 1
3 1
4 1
4 1
4 1
5 1
5 0
5 1
6 0
6 0
6 0
7 0
7 1
7 1
8 1
8 0
8 1
9 1
9 0
9 0
9 0
10 0
10 0
10 0
11 1
11 1
11 1
12 0
12 0
12 0
12 0
13 0
13 0
13 1
13 0
13 0
14 0
14 0
14 0
14 0
15 1
15 0
15 0
16 1
16 1
17 1
17 1
18 0
18 1
18 1
17 1
19 1
20 1
20 0
20 0
20 1
21 0
21 0
21 1
21 0
22 1
22 0
22 1
22 1
23 1
23 1
24 1
24 0
25 0
25 1
25 0
26 1
26 1
27 1
27 1
Okay, assuming I understood what you meant, one approach to problem 1) should be addressed using bayes filtering.
Say event L is "a fly likes sugar", event B is "a fly is in bin B".
So what you have is:
number of flies = 84
size of each bins = (eg size of bin 1: 4)
probability that a fly likes sugar:
P(L) = flies that like sugar / total number of flies = 43/84
probability that a fly doesn't like sugar:
P(notL) = 1 - P(L) = 41/84
probability that a fly is in a given bin:
P(B) = size of the bin / sum of the sizes of all bins = 4/84 (for bin 1)
probability that a fly isn't in a given bin:
P(notB) = 1 - P(B) = 80/84 (for bin 1)
probability that a fly likes sugar, knowing that's in bin B:
P(L|B) = flies that like sugar in a bin / size of the bin
(eg for bin 1 is 2/4 = 1/2)
probability that a fly likes sugar, knowing that it's not in bin B:
P(L|notB) = (total flies that like sugar - flies that like sugar in the bin)/(size of bins - size of the bin)) = 41/80
You want to know the probability that a fly is in a given bin B knowing that likes sugar, which you can obtain with:
P(B|L) = (P(L|B) * P(B)) / (P(L|B) * P(B) + P(L|notB) * P(notB))
If you compute P(B|L) and P(B|notL) for each bin, then you know which of the bins have the highest probability of containing flies that like sugar. Then you can further study those bins.
Hope i was clear, my statistics is a bit rusty and I'm not even sure I am doing everything correctly. Take it as a hint to point you in the right direction to address the problem.
You can refer here to get more accurate reasoning and results.
As for problem 2)... I have to think about it a bit more.