How to print output on a single line? - haskell

Today I was working on this code which outputs the average value of a series of "arrays", the data is inputted in this format:
3 #Number of arrays to get the average from
2 3 0 #First array
4 5 0 #Second array
1 4 5 0 #Third array
I worked on a code that outputs the data, but realized that it prints it like this:
2 #Average (Int) of the 1st array
4 #Average (Int) of the 2nd array
3 #Average (Int) of the 3rd array
(Take into account that the 0 at the end of every array is not used when calculated the average, it only exists for means of indicating end of array)
My question is, How can I properly change my code so that I can output the data like this? :
2 4 3
Here is the code I've been working in:
sumList :: [Int] -> Int sumList [] = 0 sumList (u:v) = u + sumList v
funavg :: Int -> IO () funavg numitint = if numitint==0 then return ()
else do
arrs <- getLine
let arrnum = (map read(words arrs) :: [Int])
let total = sumList arrnum
let avg = div total ((length arrnum)-1)
print avg
funavg (numitint - 1)
main :: Prelude.IO () main = do
numits <- getLine
let numitint = read numits :: Int
funavg numitint
I've searched many documents and websites, but can't come with an ideal answer.
Using recursion is mandatory.
Any help is highly appreciated :D

print is equivalent to putStrLn . show and is provided for convenience to print a single value of any Show type.
print does not have a standard library companion which omits the newline, but putStrLn does: it's called putStr. Instead of print avg, consider
putStr $ show avg

Related

Error in reading data into 3x3 matrix in Fortran [duplicate]

I would like to read and store scientific formatted numbers from a txt file, which is formatted and the numbers are separated by tabulator.
This is what I have so far:
IMPLICIT NONE
REAL,ALLOCATABLE,DIMENSION(2) :: data(:,:)
INTEGER :: row,column
INTEGER :: j,i
CHARACTER(len=30) :: filename
CHARACTER(len=30) :: format
filename='data.txt'
open(86,file=filename,err=10)
write(*,*)'open data file'
read(86, *) row
read(86, *) column
allocate(data(row,column))
format='(ES14.7)'
do i=1,row
read(86,format) data(i,:)
enddo
close(86)
This is how the txt file looks like:
200
35
2.9900E-35 2.8000E-35 2.6300E-35 2.4600E-35 2.3100E-35 2.1600E-35 ...
The problem is that it doesn't read and store the correct values from the txt to the data variable. Is it the format causing the problem?
I would also like to know how to count the number of columns in this case. (I can count the rows by using read(86,*) in a for loop.)
Yes, your format is not good for the data you show. Better one should be like that read(99,'(6(E11.4,X))') myData(i,:).
However, I am not sure if you really need to use format at your reading at all.
Following example pretty close to what you are trying to do, and it is working bot with and without format.
program readdata
implicit none
real, allocatable :: myData(:,:)
real :: myLine
integer :: i, j, myRow, myColumn
character(len=30) :: myFileName
character(len=30) :: myFormat
myFileName='data.dat'
open(99, file=myFileName)
write(*,*)'open data file'
read(99, *) myRow
read(99, *) myColumn
allocate(myData(myRow,myColumn))
do i=1,myRow
read(99,*) myData(i,:)
!read(99,'(6(E11.4,X))') myData(i,:)
print*, myData(i,:)
enddo
close(99)
end program readdata
To test, I assumed that you have rows and columns always in the file, as you give, so my test data was following.
2
6
2.9900E-35 2.8000E-35 2.6300E-35 2.4600E-35 2.3100E-35 2.1600E-35
2.9900E-35 2.8000E-35 2.6300E-35 2.4600E-35 2.3100E-35 2.1600E-35
If you are really interested to read your files with a format and if the number of columns are not constant you may need a format depending on a variable, please see related discussions here.
Though there are no direct command to count the number of items in a line, we can count the number of periods or (E|e|D|d) by using the scan command. For example,
program main
implicit none
character(100) str
integer n
read( *, "(a)" ) str
call countreal( str, n )
print *, "number of items = ", n
contains
subroutine countreal( str, num )
implicit none
character(*), intent(in) :: str
integer, intent(out) :: num
integer pos, offset
num = 0
pos = 0
do
offset = scan( str( pos + 1 : ), "." ) !! (1) search for periods
!! offset = scan( str( pos + 1 : ), "EeDd" ) !! (2) search for (E|e|D|d)
if ( offset > 0 ) then
pos = pos + offset
num = num + 1
print *, "pos=", pos, "num=", num !! just for check
else
return
endif
enddo
endsubroutine
end
Please note that pattern (1) works only when all items have periods, while pattern (2) works only when all items have exponents:
# When compiled with (1)
$ echo "2.9900 2.8000E-35 2.6300D-35 2.46 2.31" | ./a.out
pos= 2 num= 1
pos= 10 num= 2
pos= 22 num= 3
pos= 34 num= 4
pos= 40 num= 5
number of items = 5
# When compiled with (2)
$ echo "2.9900E-35 2.8000D-35 2.6300e-35 2.4600d-35" | ./a.out
pos= 7 num= 1
pos= 19 num= 2
pos= 31 num= 3
pos= 43 num= 4
number of items = 4
For more general purposes, it may be more convenient to write a custom "split()" function that separate items with white spaces (or use an external library that supports a split function).

Read scientific formatted numbers from txt

I would like to read and store scientific formatted numbers from a txt file, which is formatted and the numbers are separated by tabulator.
This is what I have so far:
IMPLICIT NONE
REAL,ALLOCATABLE,DIMENSION(2) :: data(:,:)
INTEGER :: row,column
INTEGER :: j,i
CHARACTER(len=30) :: filename
CHARACTER(len=30) :: format
filename='data.txt'
open(86,file=filename,err=10)
write(*,*)'open data file'
read(86, *) row
read(86, *) column
allocate(data(row,column))
format='(ES14.7)'
do i=1,row
read(86,format) data(i,:)
enddo
close(86)
This is how the txt file looks like:
200
35
2.9900E-35 2.8000E-35 2.6300E-35 2.4600E-35 2.3100E-35 2.1600E-35 ...
The problem is that it doesn't read and store the correct values from the txt to the data variable. Is it the format causing the problem?
I would also like to know how to count the number of columns in this case. (I can count the rows by using read(86,*) in a for loop.)
Yes, your format is not good for the data you show. Better one should be like that read(99,'(6(E11.4,X))') myData(i,:).
However, I am not sure if you really need to use format at your reading at all.
Following example pretty close to what you are trying to do, and it is working bot with and without format.
program readdata
implicit none
real, allocatable :: myData(:,:)
real :: myLine
integer :: i, j, myRow, myColumn
character(len=30) :: myFileName
character(len=30) :: myFormat
myFileName='data.dat'
open(99, file=myFileName)
write(*,*)'open data file'
read(99, *) myRow
read(99, *) myColumn
allocate(myData(myRow,myColumn))
do i=1,myRow
read(99,*) myData(i,:)
!read(99,'(6(E11.4,X))') myData(i,:)
print*, myData(i,:)
enddo
close(99)
end program readdata
To test, I assumed that you have rows and columns always in the file, as you give, so my test data was following.
2
6
2.9900E-35 2.8000E-35 2.6300E-35 2.4600E-35 2.3100E-35 2.1600E-35
2.9900E-35 2.8000E-35 2.6300E-35 2.4600E-35 2.3100E-35 2.1600E-35
If you are really interested to read your files with a format and if the number of columns are not constant you may need a format depending on a variable, please see related discussions here.
Though there are no direct command to count the number of items in a line, we can count the number of periods or (E|e|D|d) by using the scan command. For example,
program main
implicit none
character(100) str
integer n
read( *, "(a)" ) str
call countreal( str, n )
print *, "number of items = ", n
contains
subroutine countreal( str, num )
implicit none
character(*), intent(in) :: str
integer, intent(out) :: num
integer pos, offset
num = 0
pos = 0
do
offset = scan( str( pos + 1 : ), "." ) !! (1) search for periods
!! offset = scan( str( pos + 1 : ), "EeDd" ) !! (2) search for (E|e|D|d)
if ( offset > 0 ) then
pos = pos + offset
num = num + 1
print *, "pos=", pos, "num=", num !! just for check
else
return
endif
enddo
endsubroutine
end
Please note that pattern (1) works only when all items have periods, while pattern (2) works only when all items have exponents:
# When compiled with (1)
$ echo "2.9900 2.8000E-35 2.6300D-35 2.46 2.31" | ./a.out
pos= 2 num= 1
pos= 10 num= 2
pos= 22 num= 3
pos= 34 num= 4
pos= 40 num= 5
number of items = 5
# When compiled with (2)
$ echo "2.9900E-35 2.8000D-35 2.6300e-35 2.4600d-35" | ./a.out
pos= 7 num= 1
pos= 19 num= 2
pos= 31 num= 3
pos= 43 num= 4
number of items = 4
For more general purposes, it may be more convenient to write a custom "split()" function that separate items with white spaces (or use an external library that supports a split function).

Given two strings, how do I find number of reoccurences of one in another?

For example, s1='abc', s2='kokoabckokabckoab'.
Output should be 3. (number of times s1 appears in s2).
Not allowed to use for or strfind. Can only use reshape,repmat,size.
I thought of reshaping s2, so it would contain all of the possible strings of 3s:
s2 =
kok
oko
koa
oab
.... etc
But I'm having troubles from here..
Assuming you have your matrix reshaped into the format you have in your post, you can replicate s1 and stack the string such that it has as many rows as there are in the reshaped s2 matrix, then do an equality operator. Rows that consist of all 1s means that we have found a match and so you would simply search for those rows where the total sum is equal to the total length of s1. Referring back to my post on dividing up a string into overlapping substrings, we can decompose your string into what you have posted in your question like so:
%// Define s1 and s2 here
s1 = 'abc';
len = length(s1);
s2 = 'kokoabckokabckoab';
%// Hankel starts here
c = (1 : len).';
r = (len : length(s2)).';
nr = length(r);
nc = length(c);
x = [ c; r((2:nr)') ]; %-- build vector of user data
cidx = (1:nc)';
ridx = 0:(nr-1);
H = cidx(:,ones(nr,1)) + ridx(ones(nc,1),:); % Hankel subscripts
ind = x(H); % actual data
%// End Hankel script
%// Now get our data
subseqs = s2(ind.');
%// Case where string length is 1
if len == 1
subseqs = subseqs.';
end
subseqs contains the matrix of overlapping characters that you have alluded to in your post. You've noticed a small bug where if the length of the string is 1, then the algorithm won't work. You need to make sure that the reshaped substring matrix consists of a single column vector. If we ran the above code without checking the length of s1, we would get a row vector, and so simply transpose the result if this is the case.
Now, simply replicate s1 for as many times as we have rows in subseqs so that all of these strings get stacked into a 2D matrix. After, do an equality operator.
eqs = subseqs == repmat(s1, size(subseqs,1), 1);
Now, find the column-wise sum and see which elements are equal to the length of your string. This will produce a single column vector where 1 indicates that we have found a match, and zero otherwise:
sum(eqs, 2) == len
ans =
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
Finally, to add up how many times the substring matched, you just have to add up all elements in this vector:
out = sum(sum(eqs, 2) == len)
out =
2
As such, we have two instances where abc is found in your string.
Here is another one,
s1='abc';
s2='bkcokbacaabcsoabckokabckoabc';
[a,b] = ismember(s2,s1);
b = [0 0 b 0 0];
a1=circshift(b,[0 -1]);
a2=circshift(b,[0 -2]);
sum((b==1)&(a1==2)&(a2==3))
It gives 3 for your input and 4 for my example, and it seems to work well if ismember is okey.
Just for the fun of it: this can be done with nlfilter from the Image Processing Toolbox (I just discovered this function today and am eager to apply it!):
ds1 = double(s1);
ds2 = double(s2);
result = sum(nlfilter(ds2, [1 numel(ds1)], #(x) all(x==ds1)));

R: combinatorial string replacement

I am on the lookout for a gsub based function which would enable me to do combinatorial string replacement, so that if I would have an arbitrary number of string replacement rules
replrules=list("<x>"=c(3,5),"<ALK>"=c("hept","oct","non"),"<END>"=c("ane","ene"))
and a target string
string="<x>-methyl<ALK><END>"
it would give me a dataframe with the final string name and the substitutions that were made as in
name x ALK END
3-methylheptane 3 hept ane
5-methylheptane 5 hept ane
3-methyloctane 3 oct ane
5-methyloctane 5 ... ...
3-methylnonane 3
5-methylnonane 5
3-methylheptene 3
5-methylheptene 5
3-methyloctene 3
5-methyloctene 5
3-methylnonene 3
5-methylnonene 5
The target string would be of arbitrary structure, e.g. it could also be string="1-<ALK>anol" or each pattern could occur several times, as in string="<ALK>anedioic acid, di<ALK>yl ester"
What would be the most elegant way to do this kind of thing in R?
How about
d <- do.call(expand.grid, replrules)
d$name <- paste0(d$'<x>', "-", "methyl", d$'<ALK>', d$'<END>')
EDIT
This seems to work (substituting each of these into the strplit)
string = "<x>-methyl<ALK><END>"
string2 = "<x>-ethyl<ALK>acosane"
string3 = "1-<ALK>anol"
Using Richards regex
d <- do.call(expand.grid, list(replrules, stringsAsFactors=FALSE))
names(d) <- gsub("<|>","",names(d))
s <- strsplit(string3, "(<|>)", perl = TRUE)[[1]]
out <- list()
for(i in s) {
out[[i]] <- ifelse (i %in% names(d), d[i], i)
}
d$name <- do.call(paste0, unlist(out, recursive=F))
EDIT
This should work for repeat items
d <- do.call(expand.grid, list(replrules, stringsAsFactors=FALSE))
names(d) <- gsub("<|>","",names(d))
string4 = "<x>-methyl<ALK><END>oate<ALK>"
s <- strsplit(string4, "(<|>)", perl = TRUE)[[1]]
out <- list()
for(i in seq_along(s)) {
out[[i]] <- ifelse (s[i] %in% names(d), d[s[i]], s[i])
}
d$name <- do.call(paste0, unlist(out, recursive=F))
Well, I'm not exactly sure we can even produce a "correct" answer to your question, but hopefully this helps give you some ideas.
Okay, so in s, I just split the string where it might be of most importance. Then g gets the first value in each element of r. Then I constructed a data frame as an example. So then dat is a one row example of how it would look.
> (s <- strsplit(string, "(?<=l|\\>)", perl = TRUE)[[1]])
# [1] "<x>" "-methyl" "<ALK>" "<END>"
> g <- sapply(replrules, "[", 1)
> dat <- data.frame(name = paste(append(g, s[2], after = 1), collapse = ""))
> dat[2:4] <- g
> names(dat)[2:4] <- sapply(strsplit(names(g), "<|>"), "[", -1)
> dat
# name x ALK END
# 1 3-methylheptane 3 hept ane

Haskell flag pattern writing function?

I really need help in writing this function in Haskell, I don't even know where to start. Here are the specs:
Define a function flagpattern that takes a positive Int value greater than or equal to five and returns a String that can be displayed as the following `flag' pattern of dimension n, e.g.
Main> putStr (flagpattern 7)
#######
## ##
# # # #
# # #
# # # #
## ##
#######
Assuming you want a "X" enclosed in 4 lines, you need to write a function that given a coordinate (x,y) returns what character should be at that position:
coordinate n x y = if i == 0 then 'X' else ' '
(This version outputs only the leftmost X'es, modify it, remember indices start with 0)
Now you want them nicely arranged in a matrix, use a list comprehension, described in the linked text.
You should start from your problem definition:
main :: IO ()
main = putStr . flagPattern $ 7
Then, you should ask yourself about how much dots flag has:
flagPattern :: Int -> String
flagPattern = magic $ [1..numberOfDots]
Then, (hard) part of magic function should decide for each dot whether it is   or #:
partOfMagic ...
| ... = "#" -- or maybe even "#\n" in some cases?
| otherwise = " "
Then, you can concatenate parts into one string and get the answer.
Start with the type signature.
flagpattern :: Int -> String
Now break the problem into subproblems. For example, suppose I told you to produce row 2 of a size 7 flag pattern. You would write:
XX XX
Or row 3 of a size 7 flag pattern would be
X X X X
So suppose we had a function that could produce a given row. Then we'd have
flagpattern :: Int -> String
flagpattern size = unlines (??? flagrow ???)
flagrow :: Int -> Int -> String
flagrow row size = ???
unlines takes a list of Strings and turns it into a single String with newlines between each element of the list. See if you can define flagrow, and get it working correctly for any given row and size. Then see if you can use flagrow to define flagpattern.

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