I'm using jupyter notebook. Consider ar = np.array([[2,3],[5,6]]).
Then evaluating print ar displays
[[2 3]
[5 6]]
while just evaluating ar displays
array([[2, 3],
[5, 6]])
My question is: 1) What command lies actually behind this evaluation in the notebook, how could I reproduce it in a normal IDLE python script?
2) What does the second evaluation mean, which is some form of elaborate priting; does is show me the type of the object + its contents? Shouldn't it actually be ndarray instead of array?
1) I think it is more complicated than a single command. I think the code that parses which "interactivity mode" (e.g. display all, just the last line [default]) to run is here. Remember that IPython is like it's own REPL shell and there are lots of different interwoven mechanisms like this one. You could trace back through their source code and try to understand all the mechanisms involved if you wanted to, but I don't think it's a one line solution.
2) This is printing the representation of ar, repr(ar), versus the readable string form of ar. See https://stackoverflow.com/a/2626364/7458681 The reason it is array and not ndarray is that it is not the type of the object that is being printed but instead the function required in order to be able to recreate the object such that eval(repr(ar)) == ar.
Related
I've written a prime-generating function generatePrimes (full code here) that takes input bound::Int64 and returns a Vector{Int64} of all primes up to bound. After the function definition, I have the following code:
println("Generating primes...")
println("Last prime: ", generatePrimes(10^7)[end])
println("Primes generated.")
which prints, unexpectedly,
Generating primes...
9999991
Primes generated.
This output misses the "Last prime: " segment of the second print statement. The output does work as expected for smaller inputs; any input at least up to 10^6, but somehow fails for 10^7. I've tried several workarounds for this (e.g. assigning the returned value or converting it to a string before calling it in a print statement, combining the print statements, et cetera) and discovered some other weird behaviour: if the "Last prime", is removed from the second print statement, for input 10^7, the last prime doesn't print at all and all I get is a blank line between the first and third print statements. These issues are probably related, and I can't seem to find anything online about why some print statements wouldn't work in Julia.
Thanks so much for any clarification!
Edit: Per DNF's suggestion, following are some reductions to this issue:
Removing the first and last print statements doesn't change anything -- a blank line is always printed in the case I outlined and each of the cases below.
println(generatePrimes(10^7)[end]) # output: empty line
Calling the function and storing the last index in a variable before calling println doesn't change anything either; the cases below work exactly the same either way.
lastPrime::Int = generatePrimes(10^7)[end]
println(lastPrime) # output: empty line
If I call the function in whatever form immediately before a println, an empty line is printed regardless of what's inside the println.
lastPrime::Int = generatePrimes(10^7)[end]
println("This doesn't print") # output: empty line
println("This does print") # output: This does print
If I call the function (or print the pre-generated-and-stored function result) inside a println, anything before the function call (that's also inside the println) isn't printed. The 9999991 and anything else there may be after the function call is printed only if there is something else inside the println before the function call.
# Example 1
println(generatePrimes(10^7)[end]) # output: empty line
# Example 2
println("This first part doesn't print", generatePrimes(10^7)[end]) # output: 9999991
# Example 3
println("This first part doesn't print", generatePrimes(10^7)[end], " prints") # output: 9999991 prints
# Example 4
println(generatePrimes(10^7)[end], "prime doesn't print") # output: prime doesn't print
I could probably list twenty different variations of this same thing, but that probably wouldn't make things any clearer. In every single case version of this issue I've seen so far, the issue only manifests if there's that function call somewhere; println prints large integers just fine. That said, please let me know if anyone feels like they need more info. Thanks so much!
Most likely you are running this code from Atom Juno which recently has some issues with buffering standard output (already reported by others and I also sometimes have this problem).
One thing you can try to do is to flush your standard output
flush(stdout)
Like with any unstable bug restarting Atom Juno also seems to help.
I had the same issue. For me, changing the terminal renderer (File -> Settings -> Packages -> julia-client -> Terminal Options) from webgl to canvas (see pic below) seems to solve the issue.
change terminal renderer
I've also encountered this problem many times. (First time, it was triggered after using the debugger. It is probably unrelated but I have been using Julia+Juno for 2 weeks prior to this issue.)
In my case, the code before the println statement needed to have multiple dictionary assignation (with new keys) in order to trigger the behavior.
I also confirmed that the same code ran in Command Prompt (with same Julia interpreter) prints fine. Any hints about how to further investigate this will be appreciated.
I temporarily solve this issue by printing to stderr, thinking that this stream has more stringent flush mechanism: println(stderr, "hello!")
I was going through swirl() again as a refresher, and I've noticed that the author of swirl says the command ?matrix is the correct form to calling for a help screen. But, when I run ?matrix(), it still works? Is there a difference between having and not having a pair of parenthesis?
It's not specific to the swirl environment (about which I was entirely unaware until 5 minutes ago) That is standard for R. The help page for the ? shortcut says:
Arguments
topic
Usually, a name or character string specifying the topic for which help is sought.
Alternatively, a function call to ask for documentation on a corresponding S4 method: see the section on S4 method documentation. The calls pkg::topic and pkg:::topic are treated specially, and look for help on topic in package pkg.
It something like the second option that is being invoked with the command:
?matrix()
Since ?? is actually a different shortcut one needs to use this code to bring up that page, just as one needs to use quoted strings for help with for, if, next or any of the other reserved words in R:
?'?' # See ?Reserved
This is not based on a "fuzzy logic" search in hte help system. Using help instead of ? gets a different response:
> help("str()")
No documentation for ‘str()’ in specified packages and libraries:
you could try ‘??str()’
You can see the full code for the ? function by typing ? at the command line, but I am just showing how it starts the language level processing of the expressions given to it:
`?`
function (e1, e2)
{
if (missing(e2)) {
type <- NULL
topicExpr <- substitute(e1)
}
#further output omitted
By running matrix and in general any_function you get the source code of it.
I am using the Kernel Density Estimator toolbox form http://www.ics.uci.edu/~ihler/code/kde.html . But I am getting the following error when I try to execute the demo files -
>> demo_kde_3
KDE Example #3 : Product sampling methods (single, anecdotal run)
Attempt to reference field of non-structure array.
Error in double (line 10)
if (npd.N > 0) d = 1; % return 1 if the density exists
Error in repmat (line 49)
nelems = prod(double(siz));
Error in kde (line 39)
if (size(ks,1) == 1) ks = repmat(ks,[size(points,1),1]); end;
Error in demo_kde_3 (line 8)
p = kde([.1,.45,.55,.8],.05); % create a mixture of 4 gaussians for
testing
Can anyone suggest what might be wrong? I am new to Matlab and having a hard time to figure out the problem.
Thank You,
Try changing your current directory away from the #kde folder; you may have to add the #kde folder to your path when you do this. For example run:
cd('c:\');
addpath('full\path\to\the\folder\#kde');
You may also need to add
addpath('full\path\to\the\folder\#kde\examples');
Then see if it works.
It looks like function repmat (a mathworks function) is picking up the #kde class's version of the double function, causing an error. Usually, only objects of the class #kde can invoke that functions which are in the #kde folder.
I rarely use the #folder form of class definitions, so I'm not completely sure of the semantics; I'm curious if this has any effect on the error.
In general, I would not recommend using the #folder class format for any development that you do. The mathworks overhauled their OO paradigm a few versions ago to a much more familiar (and useful) format. Use help classdef to see more. This #kde code seems to predate this upgrade.
MATLAB gives you the code line where the error occurs. As double and repmat belong to MATLAB, the bug probably is in kde.m line 39. Open that file in MATLAB debugger, set a breakpoint on that line (so the execution stops immediately before the execution of that specific line), and then when the code is stopped there, check the situation. Try the entire code line in console (copy-paste or type it, do not single-step, as causing an uncatched error while single-stepping ends the execution of code in debugger), it should give you an error (but doesn't stop execution). Then try pieces of the code of that code line, what works as it should and what not, eg. does the result of size(points, 1) make any sense.
However, debugging unfamiliar code is not an easy task, especially if you're a beginner in MATLAB. But if you learn and understand the essential datatypes of MATLAB (arrays, cell arrays and structs) and the different ways they can be addressed, and apply that knowledge to the situation on the line 39 of kde.m, hopefully you can fix the bug.
Repmat calls double and expects the built-in double to be called.
However I would guess that this is not part of that code:
if (npd.N > 0) d = 1; % return 1 if the density exists
So if all is correct this means that the buil-tin function double has been overloaded, and that this is the reason why the code crashes.
EDIT:
I see that #Pursuit has already addressed the issue but I will leave my answer in place as it describes the method of detection a bit more.
I don't know what is happening, but I can't seem to add a constant to a vector. For example, typing in the console c(1,2,3,4)+5 returns 15 instead of (6,7,8,9). What am I doing wrong?
Thank you for your help.
Someone.... probably you ... has redefined the "+" function. It's easy to do:
> `+` <- function(x,y) sum(x,y)
> c(1,2,3,4)+5
[1] 15
It's easy to fix, Just use rm():
> rm(`+`)
> c(1,2,3,4)+5
[1] 6 7 8 9
EDIT: The comments (which raised the alternate possibility that c had instead been redefined as sum) are prompting me to add information about how to examine and recover from the alternative possibilities. You could use two methods to determine which of the two functions in the expression c(1,2,3,4) + 5 was the culprit. One could either type their names (with the backticks enclosing +), and note whether you got the proper definition:
> `+`
function (e1, e2) .Primitive("+")
> c
function (..., recursive = FALSE) .Primitive("c")
Using rm on the culprit (the on that doesn't match above) remains the quickest solution. Using a global rm is an in-session brainwipe:
rm(list=ls())
# all user defined objects, including user-defined functions will be removed
The advice to quit and restart would not work in some situations. If you quit-with-save, the current function definitions would be preserved. If you had earlier quit-with-save from a session where the redefinition occurred, then not saving in this session would not have fixed the problem, either. The results of prior session are held in a file named ".Rdata and this file is invisible for both Mac and Windows users because the OS file viewer (Mac's Finder.app or MS's Windows Explorer) will not display file names that begin with a "dot". I suspect that Linux users get to see them by default since using ls in a Terminal session will show them. (It's easy to find ways to change that behavior in a Mac, and that is the way I run my device.) Deleting the .Rdata file is helpful in this instance, as well as in the situation where your R session crashes on startup.
I am a bit of an R novice, and I am stuck with what seems like a simple problem, yet touches pretty deep questions about how and when things get evaluated in R.
I am using Rserve quite a bit; the typical syntax to get things evaluated remotely is a bit of a pain to type repeatedly:
RSeval(connection, quote(try(command)))
So I would like a function r which does the same thing with just the call:
r(command)
My first, naive, bound to fail attempt involved:
r <- function(command) {
RSeval(c, quote(try(command)))
}
You've guessed it: this sends, literally, try(command) to my confused Rserve daemon. I want command to be partially evaluated, if that makes any sense -- i.e. replaced by what I typed as an argument, but without evaluating it locally.
I looked for solutions to this, browsed throught the documentation for quote, substitute, eval, call, etc.. but I was not able to find something that worked. Either command gets evaluated locally, or not at all.
This is not a big problem, I can type the whole damn quote(try()) thing all the time; but at this point I am mostly curious as to how to get this to work!
EDIT:
More explanations as to what I want to do.
In the text above, command is meant to be a call do a function, ideally -- i.e., not a character string. Something like a <- 3 or assign("a", 3) rather than "a<-3" or quote(a<-3).
I believe that this is part of what makes this tricky. It seems really hard to tell R not to evaluate this locally, but only send it literally. Basically I would like my function to be a bit like quote(), which does not evaluate its argument.
Some explanation about my intentions. I wish to use Rserve frequently to pass commands to a remote R daemon. The commands would be my own (or my colleagues) and the daemon protected by firewall and authentication (and would not be run as root) -- so there is no worry of malicious commands being passed.
To be perfectly honest, this is not a big issue, and I would be pretty happy to always use the RSeval(c, quote(try())). At this point I see this more like an interesting inquiry into the subleties of R :-)
You probably want to use the substitute command, it can give you the argument unevaluated that you can build into the call.
I'm not sure if I understood you correctly - would eval(parse(text = command)) do the trick? Notice that command is a character, so you can easily pass it as a function argument. If I'm getting the point...
Anyway, evaluating user-specified commands is potentially malicious, therefore not recommended. You should either install AppArmor and tweak it (which is not an easy one), or drop the whole evaluation thing...