I am writing test-cases for a problem,I want to check my test-cases with Mathematica but I am facing some problems with file input/output.
I have to take Input from a file say "Test.in",the date consists of an Integer/String in each line and the input is terminated by EOF,I have to take the input (every line,one at a time) and in each step I have to process the input and output to a file say "output.out".How can we do this in Mathematica?
PS:I am using Mathematica 7.0
ADDED:
A sample of Test.in would be like this.
You asked to read (every line,one at a time). Well, that is surely not the Mathematica way of doing things, but as you asked for it, try something along the lines of:
strInp = OpenRead ["Test.in"];
strOut = OpenWrite ["Test.out"];
While[(str=Read[strInp, Number) != EndOfFile,
out = yourprocess[str];
Write [strOut,out];
];
Close [strOut];
Close [strInp];
(* Now show the output file *)
FilePrint ["Test.out]
Edit Other answers posted better and more Mathematica-ish ways of doing this, but that generally implies NOT reading one at a time, as Mathematica favors functional, list-wide programming rather than the iterative way.
It's rather clunky to read each value one at a time, but it's natural in M- to read them all at once and then process each one.
Here's a simple infrastructure I use all the time:
(* step one: get data *)
data = Import["ideone_fM0rs.txt", "Lines"];
(* step two: ??? *)
res = {};
Module[{value, result},
value = #;
result = yourCodeHere[value];
AppendTo[res, result];
]& /# data;
(* step three: PROFIT! *)
Export["out.txt", res, "Lines"];
but see Jon McLoone on AppendTo vs Sow/Reap for large data sets: http://blog.wolfram.com/2011/12/07/10-tips-for-writing-fast-mathematica-code.
Here's a variation with Sow/Reap for the times you'd like to collect values under various tags or categories or genuses or whatever:
data = Import["ideone_fM0rs.txt", "Lines"];
res = Reap[Module[{value, result, tag},
value = #;
result = yourCodeHere[value];
tag = generateTag[value]
Sow[result, tag];
]& /# data, _, Rule][[2]];
Export["out.txt", res, "Lines"];
It's tempting to roll all that up into a single awe-inspiring one-liner, but for maintainability I like to keep it unrolled with each step made explicit.
Of course, yourCodeHere[value] could instead be many lines of well-commented code.
Note: I downloaded your data to a local file ideone_fM0rs.txt using the download link at http://ideone.com/fM0rs
That's straightforward to do, and like everything in Mathematica there is more than one way to do it. Personally, I'd use
data = ReadList["Test.in", Number, RecordLists-> True];
and then process data using Map. There's also Import and your data probably is best loaded as type Table, although you can check the full list and see what's there. You could also use Read, but you'd have to control the file open/close yourself.
For the input aspect, this might give you a start.
vals = Import["http://ideone.com/fM0rs",
"Table"] /. {aa_ /; ! NumberQ[aa] && FreeQ[aa, List], ___} :>
Sequence[] /. {} :> Sequence[]
I think others on this forum may have better ways to go about it though; I'm not much versed in the Import/Export realm.
Daniel Lichtblau
Wolfram Research
Related
I am new to Functional programming.
The challenge I have is regarding the mental map of how a binary search tree works in Haskell.
In other programs (C,C++) we have something called root. We store it in a variable. We insert elements into it and do balancing etc..
The program takes a break does other things (may be process user inputs, create threads) and then figures out it needs to insert a new element in the already created tree. It knows the root (stored as a variable) and invokes the insert function with the root and the new value.
So far so good in other languages. But how do I mimic such a thing in Haskell, i.e.
I see functions implementing converting a list to a Binary Tree, inserting a value etc.. That's all good
I want this functionality to be part of a bigger program and so i need to know what the root is so that i can use it to insert it again. Is that possible? If so how?
Note: Is it not possible at all because data structures are immutable and so we cannot use the root at all to insert something. in such a case how is the above situation handled in Haskell?
It all happens in the same way, really, except that instead of mutating the existing tree variable we derive a new tree from it and remember that new tree instead of the old one.
For example, a sketch in C++ of the process you describe might look like:
int main(void) {
Tree<string> root;
while (true) {
string next;
cin >> next;
if (next == "quit") exit(0);
root.insert(next);
doSomethingWith(root);
}
}
A variable, a read action, and loop with a mutate step. In haskell, we do the same thing, but using recursion for looping and a recursion variable instead of mutating a local.
main = loop Empty
where loop t = do
next <- getLine
when (next /= "quit") $ do
let t' = insert next t
doSomethingWith t'
loop t'
If you need doSomethingWith to be able to "mutate" t as well as read it, you can lift your program into State:
main = loop Empty
where loop t = do
next <- getLine
when (next /= "quit") $ do
loop (execState doSomethingWith (insert next t))
Writing an example with a BST would take too much time but I give you an analogous example using lists.
Let's invent a updateListN which updates the n-th element in a list.
updateListN :: Int -> a -> [a] -> [a]
updateListN i n l = take (i - 1) l ++ n : drop i l
Now for our program:
list = [1,2,3,4,5,6,7,8,9,10] -- The big data structure we might want to use multiple times
main = do
-- only for shows
print $ updateListN 3 30 list -- [1,2,30,4,5,6,7,8,9,10]
print $ updateListN 8 80 list -- [1,2,3,4,5,6,7,80,9,10]
-- now some illustrative complicated processing
let list' = foldr (\i l -> updateListN i (i*10) l) list list
-- list' = [10,20,30,40,50,60,70,80,90,100]
-- Our crazily complicated illustrative algorithm still needs `list`
print $ zipWith (-) list' list
-- [9,18,27,36,45,54,63,72,81,90]
See how we "updated" list but it was still available? Most data structures in Haskell are persistent, so updates are non-destructive. As long as we have a reference of the old data around we can use it.
As for your comment:
My program is trying the following a) Convert a list to a Binary Search Tree b) do some I/O operation c) Ask for a user input to insert a new value in the created Binary Search Tree d) Insert it into the already created list. This is what the program intends to do. Not sure how to get this done in Haskell (or) is am i stuck in the old mindset. Any ideas/hints welcome.
We can sketch a program:
data BST
readInt :: IO Int; readInt = undefined
toBST :: [Int] -> BST; toBST = undefined
printBST :: BST -> IO (); printBST = undefined
loop :: [Int] -> IO ()
loop list = do
int <- readInt
let newList = int : list
let bst = toBST newList
printBST bst
loop newList
main = loop []
"do balancing" ... "It knows the root" nope. After re-balancing the root is new. The function balance_bst must return the new root.
Same in Haskell, but also with insert_bst. It too will return the new root, and you will use that new root from that point forward.
Even if the new root's value is the same, in Haskell it's a new root, since one of its children has changed.
See ''How to "think functional"'' here.
Even in C++ (or other imperative languages), it would usually be considered a poor idea to have a single global variable holding the root of the binary search tree.
Instead code that needs access to a tree should normally be parameterised on the particular tree it operates on. That's a fancy way of saying: it should be a function/method/procedure that takes the tree as an argument.
So if you're doing that, then it doesn't take much imagination to figure out how several different sections of code (or one section, on several occasions) could get access to different versions of an immutable tree. Instead of passing the same tree to each of these functions (with modifications in between), you just pass a different tree each time.
It's only a little more work to imagine what your code needs to do to "modify" an immutable tree. Obviously you won't produce a new version of the tree by directly mutating it, you'll instead produce a new value (probably by calling methods on the class implementing the tree for you, but if necessary by manually assembling new nodes yourself), and then you'll return it so your caller can pass it on - by returning it to its own caller, by giving it to another function, or even calling you again.
Putting that all together, you can have your whole program manipulate (successive versions of) this binary tree without ever having it stored in a global variable that is "the" tree. An early function (possibly even main) creates the first version of the tree, passes it to the first thing that uses it, gets back a new version of the tree and passes it to the next user, and so on. And each user of the tree can call other subfunctions as needed, with possibly many of new versions of the tree produced internally before it gets returned to the top level.
Note that I haven't actually described any special features of Haskell here. You can do all of this in just about any programming language, including C++. This is what people mean when they say that learning other types of programming makes them better programmers even in imperative languages they already knew. You can see that your habits of thought are drastically more limited than they need to be; you could not imagine how you could deal with a structure "changing" over the course of your program without having a single variable holding a structure that is mutated, when in fact that is just a small part of the tools that even C++ gives you for approaching the problem. If you can only imagine this one way of dealing with it then you'll never notice when other ways would be more helpful.
Haskell also has a variety of tools it can bring to this problem that are less common in imperative languages, such as (but not limited to):
Using the State monad to automate and hide much of the boilerplate of passing around successive versions of the tree.
Function arguments allow a function to be given an unknown "tree-consumer" function, to which it can give a tree, without any one place both having the tree and knowing which function it's passing it to.
Lazy evaluation sometimes negates the need to even have successive versions of the tree; if the modifications are expanding branches of the tree as you discover they are needed (like a move-tree for a game, say), then you could alternatively generate "the whole tree" up front even if it's infinite, and rely on lazy evaluation to limit how much work is done generating the tree to exactly the amount you need to look at.
Haskell does in fact have mutable variables, it just doesn't have functions that can access mutable variables without exposing in their type that they might have side effects. So if you really want to structure your program exactly as you would in C++ you can; it just won't really "feel like" you're writing Haskell, won't help you learn Haskell properly, and won't allow you to benefit from many of the useful features of Haskell's type system.
if I have
foo.txt:
cows = 3
sheep = 2
chicken = 14
dogs = 1
how do I write a lua script that can change the numbers or the animals or add elements to the list, I looked at the lua documentation on io (http://www.lua.org/manual/5.3/manual.html#6.8) and yet I still don't know how to do this without doing something stupid like having it rewrite the entire line each time it edits it.
As a matter of basic file IO operations (that is, not restricted to Lua), you cannot insert things into the middle of a file that already exists. You can overwrite data. You can append data. But you cannot insert data, such that previous data gets pushed farther down in the file.
Generally speaking, the way you deal with this is by keeping the file in memory, then writing the entire thing whenever the user says to save the data. While it's in memory, it's just a string. So you can do things like insert characters and so forth.
You can manufacture an insertion operation by reading all of the rest of the data in, going back to where you started to read from, write the new stuff, and then write the rest of the data you just read. But people generally don't do that unless it's absolutely necessary.
function add_or_update_data_in_file(key, new_value, filespec)
local file = assert(io.open(filespec))
local content = file:read'*a'
file:close()
content, cnt = content:gsub(
'(%f[%C]%s*'..key..'%s*=[ \t]*)[^\r\n]*',
'%1'..new_value, 1)
if cnt == 0 then
content = key..' = '..new_value..'\n'..content
end
file = assert(io.open(filespec, 'w'))
file:write(content)
file:close()
end
add_or_update_data_in_file('chicken', 13.5, 'foo.txt')
I recently came across some code that looked something like the following:
-record(my_rec, {f0, f1, f2...... f711}).
update_field({f0, Val}, R) -> R#my_rec{f0 = Val};
update_field({f1, Val}, R) -> R#my_rec{f1 = Val};
update_field({f2, Val}, R) -> R#my_rec{f2 = Val};
....
update_field({f711, Val}, R) -> R#my_rec{f711 = Val}.
generate_record_from_proplist(Props)->
lists:foldl(fun update_field/2, #my_rec{}, Props).
My question is about what actually happens to the record - lets say the record has 711 fields and I'm generating it from a proplist - since the record is immutable, we are, at least semantically, generating a new full record one every step in the foldr - making what looks like a function that would be linear in the length of the arguments, into one that is actually quadratic in the length, since there are updates corresponding to the length the record for every insert - Am I correct in this assumption,or is the compiler intelligent enough
to save me?
Records are tuples which first element contains the name of the records, and the next one the record fields.
The name of the fields is not stored, it is a facility for the compiler, and of course the programmer. I think it was introduced only to avoid errors in the field order when writting programs, and to allow tupple extension when releasing new version without rewriting all pattern matches.
Your code will make 712 copies of a 713 element tuple.
I am afraid, the compiler is not smart enough.
You can read more in this SO answer. If you have so big number of fields and you want to update it in O(1) time, you should use ETS tables.
I am working on fairly large Mathematica projects and the problem arises that I have to intermittently check numerical results but want to easily revert to having all my constructs in analytical form.
The code is fairly fluid I don't want to use scoping constructs everywhere as they add work overhead. Is there an easy way for identifying and clearing all assignments that are numerical?
EDIT: I really do know that scoping is the way to do this correctly ;-). However, for my workflow I am really just looking for a dirty trick to nix all numerical assignments after the fact instead of having the foresight to put down a Block.
If your assignments are on the top level, you can use something like this:
a = 1;
b = c;
d = 3;
e = d + b;
Cases[DownValues[In],
HoldPattern[lhs_ = rhs_?NumericQ] |
HoldPattern[(lhs_ = rhs_?NumericQ;)] :> Unset[lhs],
3]
This will work if you have a sufficient history length $HistoryLength (defaults to infinity). Note however that, in the above example, e was assigned 3+c, and 3 here was not undone. So, the problem is really ambiguous in formulation, because some numbers could make it into definitions. One way to avoid this is to use SetDelayed for assignments, rather than Set.
Another alternative would be to analyze the names in say Global' context (if that is the context where your symbols live), and then say OwnValues and DownValues of the symbols, in a fashion similar to the above, and remove definitions with purely numerical r.h.s.
But IMO neither of these approaches are robust. I'd still use scoping constructs and try to isolate numerics. One possibility is to wrap you final code in Block, and assign numerical values inside this Block. This seems a much cleaner approach. The work overhead is minimal - you just have to remember which symbols you want to assign the values to. Block will automatically ensure that outside it, the symbols will have no definitions.
EDIT
Yet another possibility is to use local rules. For example, one could define rule[a] = a->1; rule[d]=d->3 instead of the assignments above. You could then apply these rules, extracting them as say
DownValues[rule][[All, 2]], whenever you want to test with some numerical arguments.
Building on Andrew Moylan's solution, one can construct a Block like function that would takes rules:
SetAttributes[BlockRules, HoldRest]
BlockRules[rules_, expr_] :=
Block ## Append[Apply[Set, Hold#rules, {2}], Unevaluated[expr]]
You can then save your numeric rules in a variable, and use BlockRules[ savedrules, code ], or even define a function that would apply a fixed set of rules, kind of like so:
In[76]:= NumericCheck =
Function[body, BlockRules[{a -> 3, b -> 2`}, body], HoldAll];
In[78]:= a + b // NumericCheck
Out[78]= 5.
EDIT In response to Timo's comment, it might be possible to use NotebookEvaluate (new in 8) to achieve the requested effect.
SetAttributes[BlockRules, HoldRest]
BlockRules[rules_, expr_] :=
Block ## Append[Apply[Set, Hold#rules, {2}], Unevaluated[expr]]
nb = CreateDocument[{ExpressionCell[
Defer[Plot[Sin[a x], {x, 0, 2 Pi}]], "Input"],
ExpressionCell[Defer[Integrate[Sin[a x^2], {x, 0, 2 Pi}]],
"Input"]}];
BlockRules[{a -> 4}, NotebookEvaluate[nb, InsertResults -> "True"];]
As the result of this evaluation you get a notebook with your commands evaluated when a was locally set to 4. In order to take it further, you would have to take the notebook
with your code, open a new notebook, evaluate Notebooks[] to identify the notebook of interest and then do :
BlockRules[variablerules,
NotebookEvaluate[NotebookPut[NotebookGet[nbobj]],
InsertResults -> "True"]]
I hope you can make this idea work.
This is a saga which began with the problem of how to do survey weighting. Now that I appear to be doing that correctly, I have hit a bit of a wall (see previous post for details on the import process and where the strata variable came from):
> require(foreign)
> ipums <- read.dta('/path/to/data.dta')
> require(survey)
> ipums.design <- svydesign(id=~serial, strata=~strata, data=ipums, weights=perwt)
Error in if (nbins > .Machine$integer.max) stop("attempt to make a table with >= 2^31 elements") :
missing value where TRUE/FALSE needed
In addition: Warning messages:
1: In pd * (as.integer(cat) - 1L) : NAs produced by integer overflow
2: In pd * nl : NAs produced by integer overflow
> traceback()
9: tabulate(bin, pd)
8: as.vector(data)
7: array(tabulate(bin, pd), dims, dimnames = dn)
6: table(ids[, 1], strata[, 1])
5: inherits(x, "data.frame")
4: is.data.frame(x)
3: rowSums(table(ids[, 1], strata[, 1]) > 0)
2: svydesign.default(id = ~serial, weights = ~perwt, strata = ~strata,
data = ipums)
1: svydesign(id = ~serial, weights = ~perwt, strata = ~strata, data = ipums)
This error seems to come from the tabulate function, which I hoped would be straightforward enough to circumvent, first by changing .Machine$integer.max
> .Machine$integer.max <- 2^40
and when that didn't work the whole source code of tabulate:
> tabulate <- function(bin, nbins = max(1L, bin, na.rm=TRUE))
{
if(!is.numeric(bin) && !is.factor(bin))
stop("'bin' must be numeric or a factor")
#if (nbins > .Machine$integer.max)
if (nbins > 2^40) #replacement line
stop("attempt to make a table with >= 2^31 elements")
.C("R_tabulate",
as.integer(bin),
as.integer(length(bin)),
as.integer(nbins),
ans = integer(nbins),
NAOK = TRUE,
PACKAGE="base")$ans
}
Neither circumvented the problem. Apparently this is one reason why the ff package was created, but what worries me is the extent to which this is a problem I cannot avoid in R. This post seems to indicate that even if I were to use a package that would avoid this problem, I would only be able to access 2^31 elements at a time. My hope was to use sql (either sqlite or postgresql) to get around the memory problems, but I'm afraid I'll spend a while getting that to work, only to run into the same fundamental limit.
Attempting to switch back to Stata doesn't solve the problem either. Again see the previous post for how I use svyset, but the calculation I would like to run causes Stata to hang:
svy: mean age, over(strata)
Whether throwing more memory at it will solve the problem I don't know. I run R on my desktop which has 16 gigs, and I use Stata through a Windows server, currently setting memory allocation to 2000MB, but I could theoretically experiment with increasing that.
So in sum:
Is this a hard limit in R?
Would sql solve my R problems?
If I split it up into many separate files would that fix it (a lot of work...)?
Would throwing a lot of memory at Stata do it?
Am I seriously barking up the wrong tree somehow?
Yes, R uses 32-bit indexes for vectors so they can contain no more than 2^31-1 entries and you are trying to create something with 2^40. There is talk of introducing 64-bit indexes but that will be some way off before appearing in R. Vectors have the stated hard limit and that is it as far as base R is concerned.
I am unfamiliar with the details of what you are doing to offer any further advice on the other parts of your Q.
Why do you want to work with the full data set? Wouldn't a smaller sample that can fit in to the restrictions R places on you be just as useful? You could use SQL to store all the data and query it from R to return a random subset of more appropriate size.
Since this question was asked some time ago, I'd like to point that my answer here uses the version 3.3 of the survey package.
If you check the code of svydesign, you can see that the function that causes all the problem is within a check step that looks whether you should set the nest parameter to TRUE or not. This step can be disabled setting the option check.strata=FALSE.
Of course, you shouldn't disable a check step unless you know what you are doing. In this case, you should be able to decide yourself whether you need to set the nest option to TRUE or FALSE. nest should be set to TRUE when the same PSU (cluster) id is recycled in different strata.
Concretely for the IPUMS dataset, since you are using the serial variable for cluster identification and serial is unique for each household in a given sample, you may want to set nest to FALSE.
So, your survey design line would be:
ipums.design <- svydesign(id=~serial, strata=~strata, data=ipums, weights=perwt, check.strata=FALSE, nest=FALSE)
Extra advice: even after circumventing this problem you will find that the code is pretty slow unless you remap strata to a range from 1 to length(unique(ipums$strata)):
ipums$strata <- match(ipums$strata,unique(ipums$strata))
Both #Gavin and #Martin deserve credit for this answer, or at least leading me in the right direction. I'm mostly answering it separately to make it easier to read.
In the order I asked:
Yes 2^31 is a hard limit in R, though it seems to matter what type it is (which is a bit strange given it is the length of the vector, rather than the amount of memory (which I have plenty of) which is the stated problem. Do not convert strata or id variables to factors, that will just fix their length and nullify the effects of subsetting (which is the way to get around this problem).
sql could probably help, provided I learn how to use it correctly. I did the following test:
library(multicore) # make svy fast!
ri.ny <- subset(ipums, statefips_num %in% c(36, 44))
ri.ny.design <- svydesign(id=~serial, weights=~perwt, strata=~strata, data=ri.ny)
svyby(~incwage, ~strata, ri.ny.design, svymean, data=ri.ny, na.rm=TRUE, multicore=TRUE)
ri <- subset(ri.ny, statefips_num==44)
ri.design <- svydesign(id=~serial, weights=~perwt, strata=~strata, data=ri)
ri.mean <- svymean(~incwage, ri.design, data=ri, na.rm=TRUE)
ny <- subset(ri.ny, statefips_num==36)
ny.design <- svydesign(id=~serial, weights=~perwt, strata=~strata, data=ny)
ny.mean <- svymean(~incwage, ny.design, data=ny, na.rm=TRUE, multicore=TRUE)
And found the means to be the same, which seems like a reasonable test.
So: in theory, provided I can split up the calculation by either using plyr or sql, the results should still be fine.
See 2.
Throwing a lot of memory at Stata definitely helps, but now I'm running into annoying formatting issues. I seem to be able to perform most of the calculation I want (much quicker and with more stability as well) but I can't figure out how to get it into the form I want. Will probably ask a separate question on this. I think the short version here is that for big survey data, Stata is much better out of the box.
In many ways yes. Trying to do analysis with data this big is not something I should have taken on lightly, and I'm far from figuring it out even now. I was using the svydesign function correctly, but I didn't really know what's going on. I have a (very slightly) better grasp now, and it's heartening to know I was generally correct about how to solve the problem. #Gavin's general suggestion of trying out small data with external results to compare to is invaluable, something I should have started ages ago. Many thanks to both #Gavin and #Martin.