I'm trying to read a csv file with comments and empty lines, from which I need to fetch lines which are not empty or commented.
File looks like this:
Test File for dry run:
#This is a comment
# This is a comment with, comma
# This,is,a,comment with exact number of commas as valid lines
h1,h2,h3,h4
a,b,c,d
e,f,g,h
i,j,k,l
m,n,o,p
Expected Output:
h1 h2 h3 h4
-----------
a b c d
e f g h
i j k l
m n o p
Unsuccessful attempt:
q)("SSSS";enlist ",")0: ssr[;;]each read0 `:test.csv // tried various options with ssr but since '*' wildcard gives error with ssr so not sure of how to use regex here
This provides the required result:
q)("SSSS";enlist",")0:t where not""~/:t:5_read0`:test.csv
h1 h2 h3 h4
-----------
a b c d
e f g h
i j k l
m n o p
To ignore any number of comments you could use:
q)("SSSS";enlist",")0:t where not any each(" ";"#")~\:/:first each t:read0`:test.csv
h1 h2 h3 h4
-----------
a b c d
e f g h
i j k l
m n o p
Related
How I can remove all spaces and & from text?
col1
H&M
H & M
H & M
H &M
H& M
output
col1
HM
HM
HM
HM
HM
Help me to fix the following code or give me new one:
df.withColumn('col1', F.regexp_replace("col1", "&", ""))
Also how I can get '' if there is space between characters and words and ' ' if there is no white space between character and words.
col1
H&M
H & M
H & M
H &M
H& M
output
col1
H M
H M
H M
H M
H M
Replace all characters except alphabets
df = df.withColumn('col1',regexp_replace('col1','[^A-Z]','')).show()
{a^p b^p; p is a prime number}
{a^p b^p; p is a prime number, m is a fixed number and m≥p≥0}
How do I prove if this is a regular language/context free language (or not)?
1) L = {a^n b^n; n is a prime number} :
So the prove can be done by contradiction. Suppose L is regular, and p is the pumping length.
The test string is w = a^p b^p, w belongs to L, and |w| = 2p >= p
We subdivide w=xyz. There are 3 conditions to prove the pumping lemma:
from the third condition, |xy| < p, so xy contains only a's
from the second condition, |y| > 0, so y has the form y = a^k, where 1 <= k <= p
from the first condition, xy^iz belongs to L for i = 0, 1, 2, ... So if you pump down (i = 0) you got:
w = a^(p - k) b^p , and w does not belongs to L (Because the quantity of a's and b's are different)
So you prove that L is not regular.
I have two 1x6 vectors that I am eventually trying to just sum up, but I need to get all of the possible combinations of these vectors before doing so. The vectors will look like so:
V1=[a b c d e f];
V2=[A B C D E F];
What I need is to find all possible combinations of variables that will remain a 1x6 vector. I have been messing around for a while now and I think I have found a way by using various matrices but it seems terribly inefficient. An example of what I am looking for is as follows.
M=[a b c d e f;
A b c d e f;
A B c d e f;
A B C d e f;
A B C D e f;
A B C D E f;
A B C D E F;
. . .]
And so on and so forth until all combinations are found. Unfortunately I am not a MATLAB whiz hence the reason I'm reaching out. I'm sure there has to be a much simpler way than what I have been trying. I hope that my question was relatively clear. Any help is much appreciated! Thanks!
I used cellfun to create the indexes:
V1=['abcdef'];
V2=['ABCDEF'];
VV = [V1;V2];
l = length(V1);
pows = 0:l-1;
x = num2cell(2.^pows);
L = x{end};
rows = cellfun(#(x) reshape([ones(x,L/x);2*ones(x,L/x)],[2*L 1]),x,'Uniformoutput',0);
rows = cell2mat(rows);
cols = repmat(1:l,[2*L 1]);
idxs = sub2ind(size(VV),rows,cols);
M = VV(idxs);
and you get:
M =
abcdef
Abcdef
aBcdef
ABcdef
abCdef
AbCdef
aBCdef
ABCdef
abcDef
AbcDef
...
This is a very simple question, but found surprisingly very little about it online...
I want to find the minimizer of a function in maple, I am not sure how to indicate which is the variable of interest? Let us take a very simple case, I want the symbolic minimizer of a quadratic expression in x, with parameters a, b and c.
Without specifying something, it does minimize over all variables, a, b, c and x.
f4 := a+b*x+c*x^2
minimize(f4, location)
I tried to specify the variable in the function, did not work either:
f5 :=(x) ->a+b*x+c*x^2
minimize(f5, location)
How should I do this? And, how would I do if I wanted over two variables, x and y?
fxy := a+b*x+c*x^2 + d*y^2 +e*y
f4 := a+b*x+c*x^2:
extrema(f4, {}, x);
/ 2\
|4 a c - b |
< ---------- >
| 4 c |
\ /
fxy := a+b*x+c*x^2 + d*y^2 +e*y:
extrema(fxy, {}, {x,y});
/ 2 2\
|4 a c d - b d - c e |
< --------------------- >
| 4 c d |
\ /
The nature of the extrema will depend upon the values of the parameters. For your first example above (quadratic in x) it will depend on the signum of c.
The command extrema accepts an optional fourth argument, such as an unassigned name (or an uneval-quoted name) to which is assigns the candidate solution points (as a side-effect of its calculation). Eg,
restart;
f4 := a+b*x+c*x^2:
extrema(f4, {}, x, 'cand');
2
4 a c - b
{----------}
4 c
cand;
b
{{x = - ---}}
2 c
fxy := a+b*x+c*x^2 + d*y^2 +e*y:
extrema(fxy, {}, {x,y}, 'cand');
2 2
4 a c d - b d - c e
{---------------------}
4 c d
cand;
b e
{{x = - ---, y = - ---}}
2 c 2 d
Alternatively, you may set up the partial derivatives and solve them manually. Note that for these two examples there is just a one result (for each) returned by solve.
restart:
f4 := a+b*x+c*x^2:
solve({diff(f4,x)},{x});
b
{x = - ---}
2 c
normal(eval(f4,%));
2
4 a c - b
----------
4 c
fxy := a+b*x+c*x^2 + d*y^2 +e*y:
solve({diff(fxy,x),diff(fxy,y)},{x,y});
b e
{x = - ---, y = - ---}
2 c 2 d
normal(eval(fxy,%));
2 2
4 a c d - b d - c e
---------------------
4 c d
The code for the extrema command can be viewed, by issuing the command showstat(extrema). You can see how it accounts for the case of solve returning multiple results.
I'm using Ruby to read and then print a file to stdout, redirecting the output to a file in Windows PowerShell.
However, when I inspect the files, I get this for the input:
PS D:> head -n 1 .\inputfile
<text id="http://observer.guardian.co.uk/osm/story/0,,1009777,00.html"> <s> Hooligans NNS hooligan
, , , unbridled JJ unbridled passion NN passion
- : - and CC and no DT no executive JJ executiv
e boxes NNS box . SENT . </s>
... yet this for the output:
PS D:> head -n 1 .\outputfile
ÿ_< t e x t i d = " h t t p : / / o b s e r v e r . g u a r d i a n . c o . u k / o s m / s t o r y / 0 , , 1 0 0 9 7 7 7 , 0
0 . h t m l " > < s > H o o l i g a n s N N S h o o l i g a n , ,
, u n b r i d l e d J J u n b r i d l e d p a s s i o n N N p a s s i o n
- : - a n d C C a n d n o D T n o e x e c u t i v e J J
e x e c u t i v e b o x e s N N S b o x . S E N T . < / s >
How can this happen?
Edit: since my problem didn't have anything to do with Ruby, I've removed the Ruby-code, and included my usage of the Windows shell.
In PowerShell > is effectively the same as | Out-File and Out-File defaults to Unicode encoding. Try this instead of using >:
... | Out-File outputfile -encoding ASCII