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
Related
I have the following minimal example data (in reality 100's of groups) in range A1:P9 (same data in range A14:A22):
With Sample A1:AR9:
2
61
219
2
4
2
:
61
219
26
26
26
94
21
33
4
26
26
26
94
2
2
:
154
26
40
19
3
2
21
33
14
1
2
3
:
87
39
54
38
26
32
38
26
32
87
39
54
38
26
23
23
4
6
28
2
154
26
2
2
40
19
14
87
39
54
38
26
32
38
26
32
87
39
54
38
26
1
23
2
23
4
4
3
6
20
28
Or Sample A14:AQ22:
2
61
219
2
:
61
219
4
:
26
26
26
94
2
:
21
33
4
26
26
26
94
2
:
154
26
2
:
40
19
3
2
21
33
14
:
87
39
54
38
26
32
38
26
32
87
39
54
38
26
1
:
23
2
:
23
4
:
3
6
20
2
154
26
2
2
40
19
14
87
39
54
38
26
32
38
26
32
87
39
54
38
26
1
23
2
23
4
4
3
6
20
28
I need the output as shown in range Q1:AR3 or as in range Q14:AQ16.
Basically, at each group delimited/inbetween values in Column A, I would need:
The intemediary adjacent values in Column B to be transposed horizontally
And the adjacent content of Columns C to P (14 Columns, at least) to be "joined" together horizontaly an sequencialy "per group", including the content of the delimiter's row (in Column A).
As a bonus it would be really nice to have the Transposed data followed by a :, and each sub Content of Columns C to P to be also separated by a | (as shown in screenshot Q1:AR3 or Q14:AR16).
(Or if it's more feasible, alternatively, the simpler to read 2nd model as in A14:AQ22).
I have a really hard time putting together a formula to come to the expected result.
All I could think of was:
Transposing Column B's content by getting the rows of the adjacent Cells with values in column A,
Concatenating with the Column letter,
Duplicating it in a new column, and Filtering out the blank intermediary cells,
Then shifting the duplicated column 1 cell up,
Then concatenating within a TRANSPOSE formula to get the range of the groups,
Then finally transposing all the groups from Columns B in a new Colum
(very convoluted but I couldn't find better way).
To get to that input:
=TRANSPOSE(B1:B3)
=TRANSPOSE(B4:B5)
=TRANSPOSE(B7:B9)
That was already a very manual and error prone process, and still I could not successfully think of how to do the remaining content joining of Column C to P in a formula.
I tested the following approach but it's not working and would be very tedious process to fix to go and to implement on large datasets:
=TRANSPOSE(B1:B3)&": "&JOIN( " | " , FILTER(C1:P1, NOT(C2:P2 = "") ))&JOIN( " | " , FILTER(C2:P2, NOT(C2:P2 = "") ))&JOIN( " | " , FILTER(C43:P3, NOT(C3:P3 = "") ))
=TRANSPOSE(B4:B5)&": "&JOIN( " | " , FILTER(C4:P4, NOT(C4:P4 = "") ))&JOIN( " | " , FILTER(C5:P5, NOT(C5:P5 = "") ))
=TRANSPOSE(B6:B9)&": "&JOIN( " | " , FILTER(C6:P6, NOT(C6:P6 = "") ))&JOIN( " | " , FILTER(C7:P7, NOT(C7:P7 = "") ))&JOIN( " | " , FILTER(C8:P8, NOT(C8:P8 = "") ))&JOIN( " | " , FILTER(C8:P8, NOT(C9:P9 = "") ))
What better approach to favor toward the expected result? Preferably with a Formula, or if not possible with a script.
Any help is greatly appreciated.
For Sample 1 try this out:
=LAMBDA(norm,MAP(UNIQUE(norm),LAMBDA(ζ,{TRANSPOSE(FILTER(B1:B9,norm=ζ)),":",SPLIT(BYROW(TRANSPOSE(FILTER(BYROW(C1:P9,LAMBDA(r,TEXTJOIN("ζ",1,r))),norm=ζ)),LAMBDA(rr,TEXTJOIN("γ|γ",1,rr))),"ζγ")})))(SORT(SCAN(,SORT(A1:A9,ROW(A1:A9),),LAMBDA(a,c,IF(c="",a,c))),ROW(A1:A9),))
Anyone got any ideas on how to do this?
I'm trying to build a spreadsheet that helps me monitor the performance of my blog articles. So if the article historically had >=100 visits at any point but subsequently gets <100 at any point I want to know about it.
The formula I've been playing with is:
=IF(((FILTER(C2:G2,C2:G2<>E2))>=100 AND (FILTER(C2:G2,C2:G2<>E2))<100, "Article Failing", ""))
I'm using Filter btw because I need to exclude column E, which is the delta between this month's & last month's numbers.
I know the formula isn't logically right but struggling to think of a way to do it.
Edit:
Here's a link to the spreadsheet with desired output https://docs.google.com/spreadsheets/d/1TeaQ6oUbJDeKxUi8tvvCWXtw0oK9d5IVO60j1UbQCK8/edit?usp=sharing
Here's a table showing the sample data and desired output:
Total users (last 30 days)
Total users (prev 30 days)
Delta - Total users
Total users last 30-60 days
Total users prev 60-90 days
Delta - Total users
Above 100
Article Failing
651
90
-417
772
249
523
Tweak Article
Failing
610
570
40
550
432
118
Tweak Article
OK
436
409
27
328
210
118
Tweak Article
OK
422
288
134
53
288
-235
Tweak Article
OK
95
476
-90
417
477
-60
Below100
Failing
337
179
158
129
182
-53
Tweak Article
OK
305
395
-90
318
343
-25
Tweak Article
OK
304
348
-44
299
253
46
Tweak Article
OK
302
277
25
283
317
-34
Tweak Article
OK
286
252
34
268
281
-13
Tweak Article
OK
213
193
20
221
168
53
Tweak Article
OK
157
138
19
132
166
-34
Tweak Article
OK
150
157
-7
110
68
42
Tweak Article
OK
I've made cells B2 & A6 be failing articles i.e. they were >=100 but have since gone below 100. The end column 'Article Failing' is where I'm trying to create the formula.
Hope that makes things a bit clearer.
This formula will match the desired results you show in the sample spreadsheet:
=if(
(max(A$2:A2) >= 100) * (A2 < 100)
+
(max(B$2:B2) >= 100) * (B2 < 100)
+
(row(B2) = row(B$2)) * (B2 < 100),
"Failing",
"OK"
)
I basically want the same thing as this OP:
Is there a J idiom for adding to a list until a certain condition is met?
But I cant get the answers to work with OP's function or my own.
I will rephrase the question and write about the answers at the bottom.
I am trying to create a function that will return a list of fibonacci numbers less than 2.000.000. (without writing "while" inside the function).
Here is what i have tried:
First, i picked a way to culculate fibonacci numbers from this site:
https://code.jsoftware.com/wiki/Essays/Fibonacci_Sequence
fib =: (i. +/ .! i.#-)"0
echo fib i.10
0 1 1 2 3 5 8 13 21 34
Then I made an arbitrary list I knew was larger than what I needed. :
fiblist =: (fib i.40) NB. THIS IS A BAD SOLUTION!
Finally, I removed the numbers that were greater than what I needed:
result =: (fiblist < 2e6) # fiblist
echo result
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1.34627e6
This gets the right result, but is there a way to avoid using some arbitrary number like
40 in "fib i.40" ?
I would like to write a function, such that "func 2e6" returns the list of fibonacci numbers below 2.000.000. (without writing "while" inside the function).
echo func 2e6
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1.34627e6
here are the answers from the other question:
first answer:
2 *^:(100&>#:])^:_"0 (1 3 5 7 9 11)
128 192 160 112 144 176
second answer:
+:^:(100&>)^:(<_) ] 3
3 6 12 24 48 96 192
As I understand it, I just need to replace the functions used in the answers, but i dont see how
that can work. For example, if I try:
echo (, [: +/ _2&{.)^:(100&>#:])^:_ i.2
I get an error.
I approached it this way. First I want to have a way of generating the nth Fibonacci number, and I used f0b from your link to the Jsoftware Essays.
f0b=: (-&2 +&$: -&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.
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.
I try to implement a simple ARMA model, however have serious difficulties getting it to run. When adding a parameter to the error term everything works fine (see the return x_m1 + a*e statement, commented out below), however if I add a parameter to the auto regressive part, I get a FloatingPointError or LinAlgError or PositiveDefiniteError, depending on the initialization method I use.
The code is also put into a gist you can find here. The model definition is replicated here:
with pm.Model() as model:
a = pm.Normal("a", 0, 1)
sigma = pm.Exponential('sigma', 0.1, testval=F(.1))
e = pm.Normal("e", 0, sigma, shape=(N-1,))
def x(e, x_m1, a):
# return x_m1 + a*e
return a*x_m1 + e
x, updates = theano.scan(
fn=x,
sequences=[e],
outputs_info=[tt.as_tensor_variable(data.iloc[0])],
non_sequences=[a]
)
x = pm.Deterministic('x', x)
lam = pm.Exponential('lambda', 5.0, testval=F(.1))
y = pm.StudentT("y", mu=x, lam=lam, nu=1, observed=data.values[1:]) #
with model:
trace = pm.sample(2000, init="NUTS", n_init=1000)
Here the errors respective to the initialization methods:
"ADVI" / "ADVI_MAP": FloatingPointError: NaN occurred in ADVI optimization.
"MAP": LinAlgError: 35-th leading minor not positive definite
"NUTS": PositiveDefiniteError: Scaling is not positive definite. Simple check failed. Diagonal contains negatives. Check indexes [ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71]
For details on the error messages, please look at this github issue posted at pymc3.
To be explicit, I really would like to have a scan-like solution which is easily extendable to for instance a full ARMA model. I know that one can represent the presented AR(1) model without scan by defining logP as already done in pymc3/distributions/timeseries.py#L18-L46, however I was not able to extend this vectorized style to a full ARMA model. The use of theano.scan seems preferable I think.
Any help is highly appriciated!