I am performing a mathematical function on a list of numbers, however I'm getting the result as a generator instead of another list.
I have tried putting the 'for' loop first but it gives me typeMonoidElement
ciphertext = ([((G**block)*([choice(list(range(n)))]**n))%(n**2)] for block in blocks)
I expected the output to be a list of numbers but im getting a generator instead.
I don't know what block is, but in general a lot of Python stuff does now return generators instead of lists. I just made up something to complete your example.
blocks = [1,2,24,5]
ciphertext = ([((G**block)*([choice(list(range(n)))]**n))%(n**2)] for block in blocks)
cipher text
This returns <generator object <genexpr> at 0x7fcf3c5403c0> or the like. Actually, this is a feature! Now you don't have to hold the entire list in memory, and can iterate over it whenever you find it convenient.
However, if you want a list, you can do it in two ways.
list(cipher text)
[block^2 for block in blocks]
You used parentheses rather than brackets (in American English usage); the parens makes a generator. Hope this helps.
On a separate note, your code needs a lot of extra stuff to work. The generator I just created doesn't notice that G is undefined, and that n is currently a function, not an integer! This is a simpler example of the same thing.
blocks = [1,2,24,5]
ciphertext = (block^2 for block in blocks)
print ciphertext
list(cipher text)
<generator object <genexpr> at 0x7fcf3c540410>
[1, 4, 576, 25]
Also note you are allowed in Sage to do x^2 rather than x**2. Enjoy Sage!
Related
I am new to Julia so sorry if this question is obvious.
I am trying to use Julia to help me run a series of finite element models, which use a text input file to give instructions to the finite element solver. Basically, I would like to use Julia to read in the base input file, edit some parameters on some lines of the file and then write it as a new file. I am getting hung up on a couple things though.
Currently, I am reading in the file like this
mdl = "fullmodelSVTV"; #name of input file
A = readlines(mdl*".inp")
This read each line from the file in as a separate string in a vector which I like because it makes it easier to edit the sections I want but it also makes things more difficult when I try to write to a new file.
I am writing the file like this.
io = open("name.inp","w")
print(io,A)
close(io)
When I try to write to a new file the output ends up look like this
Output from code
which is ["string at index 1","string at index 2","string at index 3"...].
What I would like to do is output this the exact same way is it is read in with string at each index of the vector on its own line. I would also like to remove the brackets and quotation marks from the file, as they might interfere with the finite element solver.
I think I have found a way to concatenate all of the strings at each index and separated them with a new line like shown below.
for i in 1:length(A)
conc = conc*"\n"*lines[i]
end
The issue with this is that it takes a long time to do given the size of the input files I am working with and I feel like there has to achieve my goal.
I also cannot find a way to remove the brackets or quotation marks when writing the file.
So, I'm wondering if anyone has any advice for a better way to write these text files in terms of both concatenating all of the strings from the vector when outputting as well as outputting without the brackets and quotation marks.
Thanks, any advice is appreciated.
The issue with print(io,A) is that it is printing a representation of the vector, but in fact you want to print each element of the vector. To do so, you can simply print each line in a loop:
open("name.inp", "w") do io
for line in A
println(io, line)
end
end
This avoids the overhead of string concatenation.
I work with knitr() and I wish to transform inline Latex commands like "\label" and "\ref", depending on the output target (Latex or HTML).
In order to do that, I need to (programmatically) generate valid R strings that correctly represent the backslash: for example "\label" should become "\\label". The goal would be to replace all backslashes in a text fragment with double-backslashes.
but it seems that I cannot even read these strings, let alone process them: if I define:
okstr <- function(str) "do something"
then when I call
okstr("\label")
I directly get an error "unrecognized escape sequence"
(of course, as \l is faultly)
So my question is : does anybody know a way to read strings (in R), without using the escaping mechanism ?
Yes, I know I could do it manually, but that's the point: I need to do it programmatically.
There are many questions that are close to this one, and I have spent some time browsing, but I have found none that yields a workable solution for this.
Best regards.
Inside R code, you need to adhere to R’s syntactic conventions. And since \ in strings is used as an escape character, it needs to form a valid escape sequence (and \l isn’t a valid escape sequence in R).
There is simply no way around this.
But if you are reading the string from elsewhere, e.g. using readLines, scan or any of the other file reading functions, you are already getting the correct string, and no handling is necessary.
Alternatively, if you absolutely want to write LaTeX-like commands in literal strings inside R, just use a different character for \; for instance, +. Just make sure that your function correctly handles it everywhere, and that you keep a way of getting a literal + back. Here’s a suggestion:
okstr("+label{1 ++ 2}")
The implementation of okstr then needs to replace single + by \, and double ++ by + (making the above result in \label{1 + 2}). But consider in which order this needs to happen, and how you’d like to treat more complex cases; for instance, what should the following yield: okstr("1 +++label")?
I have a list of hosts inbound in the form of one string separated by commas.
EXAMPLE: "host01,host02,host03,"
I have this line that was an array of strings but I need it to be a map[string]interface{}
Here is what it is how do I make it a map[string]interface{}?
• Removing the trailing or any trailing comma.
hosts := []string{strings.TrimSuffix(hostlist, ",")}
• Later I split them on the comma like this.
hosts = strings.split(hosts[0], ",")
I just need to make it so names are keys and the values are unknown from APIs so an interface{}.
Thanks and forgive me I know this is super simple I am just not seeing it.
Loop over your slice of strings. Set each map entry to nil.
There is no fancy syntax like Python's list comprehensions or Perl's freaky group assignments.
And remember that StackOverflow's tag info is often really useful. See: https://stackoverflow.com/tags/go/info
And from there to the language specification. One bit that will help is https://golang.org/ref/spec#For_range if you aren't familiar with Go's for syntax to loop over slices.
Basically I want someone to give me a simple rundown of how this bit of python code works. Much appreciated
vari :
kw1 = ['keyword1', 'keyword2']
problem = input("Detect keywords from list\n")
main :
if set(kw1).intersection(problem.split()):
print(" Kw found. ")
else:
print(" Keywords not found. ")
A lot of things there.
First, when you call input you're asking for the user to give you an input string.
When you use split() on it you transform it into a list of strings, by separating the input string based on the empty spaces, so that "bla bli blo".split() gives you ["bla","bli","blo"].
Then, when you call set(my_list), it will transform my_list into a set, which is a mathematical construct without any duplicates and which responds to operators like union, intersection and so on.
Finally, when you compare your set (made from splitting the user input) to a list of keywords, if there are no matches (so none of the keywords in the list appreared directly in the user input), then it will give you an empty set and that will be considered as false by the if. So if set(["bla","bli","blo"]).intersection(["blu"]) will not activate, but if set(["bla","bli","blo"]).intersection(["blu","blo"]) will, as it is not an empty set.
Note that if you want to recognize keywords inside words, this method will NOT work. For instance, if you're looking for keywords kw1=['car','truck','bike'] and the user inputs cars trucks bikes, none of the keywords will be recognized, because the split() will split along empty spaces, giving you ['cars','trucks','bikes'] and 'cars'!='car'...
I'm trying to learn how to configure my .vimrc file with my own functions.
I'd like to write a function that traverses every line in a file and counts the total number of characters, but ignores all whitespace. This is for a programming exercise and as a stepping stone to more complex programs (I know there are other ways to get this example value using Vim or external programs).
Here's what I have so far:
function countchars()
let line = 0
let count = 0
while line < line("$")
" update count here, don't count whitespace
let line = getline(".")
return count
endfun
What functional code could I replace that commented line with?
If I understand the question correctly, you're looking to count the number of non-whitespace characters in a line. A fairly simple way to do this is to remove the whitespace and look at the length of the resulting line. Therefore, something like this:
function! Countchars()
let l = 1
let char_count = 0
while l <= line("$")
let char_count += len(substitute(getline(l), '\s', '', 'g'))
let l += 1
endwhile
return char_count
endfunction
The key part of the answer to the question is the use of substitute. The command is:
substitute(expr,pattern,repl,flags)
expr in this case is getline(l) where l is the number of the line being iterated over. getline() returns the content of the line, so this is what is being parsed. The pattern is the regular expression \s which matches any single whitespace character. It is replaced with '', i.e. an empty string. The flag g makes it repeat the substitute as many times as whitespace is found on the line.
Once the substitution is complete, len() gives the number of non-whitespace characters and this is added to the current value of char_count with +=.
A few things that I've changed from your sample:
The function name starts with a capital letter (this is a requirement for user defined functions: see :help user-functions)
I've renamed count to char_count as you can't have a variable with the same name as a function and count() is a built-in function
Likewise for line: I renamed this to l
The first line in a file is line 1, not line 0, so I initialised l to 1
The while loop counted up to but not including the last line, I assume you wanted all the lines in the file (this is probably related to the line numbering starting at 1): I changed your code to use <= instead of <
Blocks aren't based on indentation in vim, so the while needs an endwhile
In your function, you have let line = getline('.')
I added a ! on the function definition as it makes incremental development much easier (everytime you re-source the file, it will override the function with the new version rather than spitting out an error message about it already existing).
Incrementing through the file works slightly differently...
In your function, you had let line = getline('.'). Ignoring the variable name, there are still some problems with this implementation. I think what you meant was let l = line('.'), which gives the line number of the current line. getline('.') gives the contents of the current line, so the comparison on the while line would be comparing the content of the current line with the number of the last line and this would fail. The other problem is that you're not actually moving through the file, so the current line would be whichever line you were on when you called the function and would never change, resulting in an infinite loop. I've replaced this with a simple += 1 to step through the file.
There are ways in which the current line would be a useful way to do this, for example if you were writing a function with that took a range of lines, but I think I've written enough for now and the above will hopefully get you going for now. There are plenty of people on stackoverflow to help with any issues anyway!
Have a look at:
:help usr_41.txt
:help function-list
:help user-functions
:help substitute()
along with the :help followed by the various things I used in the function (getline(), line(), let+= etc).
Hope that was helpful.
This approach uses lists:
function! Countchars()
let n = 0
for line in getline(1,line('$'))
let n += len(split(line,'\zs\s*'))
endfor
return n
endfunction
I suppose you have already found the solution.
Just for info:
I use this to count characters without spaces in Vim:
%s/\S/&/gn