digitalmars.D.learn - Using iteration / method chaining / etc on multi-dimensional arrays

Chris Katko (17/17) Apr 12 2018 I googled but couldn't find any clear solution.

Uknown (5/22) Apr 12 2018 I think something like this would work:

Chris Katko (33/65) Apr 12 2018 But each doesn't return anything, it mutates, right? I think

Chris Katko (1/1) Apr 12 2018 Wait, that might not be the error.

Chris Katko (7/8) Apr 12 2018 Just the top one. This one:

Paul Backus (4/9) Apr 12 2018 You need to put an exclamation point after 'each' to pass the

Chris Katko (22/33) Apr 12 2018 DOH.

=?UTF-8?Q?Ali_=c3=87ehreli?= (24/66) Apr 12 2018 Assuming the file "2darray" includes

Seb (5/22) Apr 12 2018 Use mir for multi-dimensional arrays:

Chris Katko <ckatko gmail.com> writes:

I googled but couldn't find any clear solution.

I've got a 2-D array of strings read from a text file I parsed. 
So it's like

0 1 15 0 0
2 12 1 0 0
...
0 1 0 10 0

They come in with spaces, so I join into an array between them. 
But then the last ones have a newline \n on the end, which 
explodes the to! type conversion.

If it was a single array, I could simply do:

string [25] test;

   test.each((ref n) => n.stripRight());

But how do you do that when it's a 2-D array?

I'm looking not just for this case, but the generate case of 
iterating / applying filters to 2-D arrays.

Thanks.

Apr 12 2018

Uknown <sireeshkodali1 gmail.com> writes:

On Thursday, 12 April 2018 at 15:38:34 UTC, Chris Katko wrote:
 I googled but couldn't find any clear solution.

 I've got a 2-D array of strings read from a text file I parsed. 
 So it's like

 0 1 15 0 0
 2 12 1 0 0
 ...
 0 1 0 10 0

 They come in with spaces, so I join into an array between them. 
 But then the last ones have a newline \n on the end, which 
 explodes the to! type conversion.

 If it was a single array, I could simply do:

 string [25] test;

   test.each((ref n) => n.stripRight());

 But how do you do that when it's a 2-D array?

 I'm looking not just for this case, but the generate case of 
 iterating / applying filters to 2-D arrays.

 Thanks.

I think something like this would work:

test.each((ref s) => s.each((ref n) => n.stripRight()));

Essentially, get each 1D array from the 2D array, and then apply 
the algorithms on that. There's probably a better way though

Apr 12 2018

Chris Katko <ckatko gmail.com> writes:

On Thursday, 12 April 2018 at 15:47:14 UTC, Uknown wrote:
 On Thursday, 12 April 2018 at 15:38:34 UTC, Chris Katko wrote:
 I googled but couldn't find any clear solution.

 I've got a 2-D array of strings read from a text file I 
 parsed. So it's like

 0 1 15 0 0
 2 12 1 0 0
 ...
 0 1 0 10 0

 They come in with spaces, so I join into an array between 
 them. But then the last ones have a newline \n on the end, 
 which explodes the to! type conversion.

 If it was a single array, I could simply do:

 string [25] test;

   test.each((ref n) => n.stripRight());

 But how do you do that when it's a 2-D array?

 I'm looking not just for this case, but the generate case of 
 iterating / applying filters to 2-D arrays.

 Thanks.

 I think something like this would work:

 test.each((ref s) => s.each((ref n) => n.stripRight()));

 Essentially, get each 1D array from the 2D array, and then 
 apply the algorithms on that. There's probably a better way 
 though

But each doesn't return anything, it mutates, right? I think 
that's the problem I ran into with my attempt. With your code, I 
get an error about void:

   string []x = split(file.readln.idup, " ");
   x.each((ref s) => s.each((ref n) => n.stripRight()));

extra.d(2493): Error: template std.algorithm.iteration.each 
cannot deduce function from argument types !()(string[], void), 
candidates are:
/usr/include/dmd/phobos/std/algorithm/iteration.d(864):        
std.algorithm.iteration.each(alias pred = "a")

extra.d(2499): Error: template std.algorithm.mutation.stripRight 
cannot deduce function from argument types !()(string), 
candidates are:
/usr/include/dmd/phobos/std/algorithm/mutation.d(2363):        
std.algorithm.mutation.stripRight(Range, E)(Range range, E 
element) if (isBidirectionalRange!Range && is(typeof(range.back 
== element) : bool))
/usr/include/dmd/phobos/std/algorithm/mutation.d(2375):        
std.algorithm.mutation.stripRight(alias pred, Range)(Range range) 
if (isBidirectionalRange!Range && is(typeof(pred(range.back)) : 
bool))

extra.d(2500): Error: template std.algorithm.mutation.stripRight 
cannot deduce function from argument types !()(string), 
candidates are:
/usr/include/dmd/phobos/std/algorithm/mutation.d(2363):        
std.algorithm.mutation.stripRight(Range, E)(Range range, E 
element) if (isBidirectionalRange!Range && is(typeof(range.back 
== element) : bool))
/usr/include/dmd/phobos/std/algorithm/mutation.d(2375):        
std.algorithm.mutation.stripRight(alias pred, Range)(Range range) 
if (isBidirectionalRange!Range && is(typeof(pred(range.back)) : 
bool))

Apr 12 2018

Chris Katko <ckatko gmail.com> writes:

Wait, that might not be the error.

Apr 12 2018

Chris Katko <ckatko gmail.com> writes:

On Thursday, 12 April 2018 at 20:37:49 UTC, Chris Katko wrote:
 Wait, that might not be the error.

Just the top one. This one:

extra.d(2493): Error: template std.algorithm.iteration.each 
cannot deduce function from argument types !()(string[], void), 
candidates are:
/usr/include/dmd/phobos/std/algorithm/iteration.d(864):        
std.algorithm.iteration.each(alias pred = "a")

Apr 12 2018

Paul Backus <snarwin gmail.com> writes:

On Thursday, 12 April 2018 at 20:34:40 UTC, Chris Katko wrote:
 But each doesn't return anything, it mutates, right? I think 
 that's the problem I ran into with my attempt. With your code, 
 I get an error about void:

   string []x = split(file.readln.idup, " ");
   x.each((ref s) => s.each((ref n) => n.stripRight()));

You need to put an exclamation point after 'each' to pass the 
function as a template parameter:

     x.each!((ref s) => s.each!((ref n) => n.stripRight()));

Apr 12 2018

Chris Katko <ckatko gmail.com> writes:

On Thursday, 12 April 2018 at 21:17:30 UTC, Paul Backus wrote:
 On Thursday, 12 April 2018 at 20:34:40 UTC, Chris Katko wrote:
 But each doesn't return anything, it mutates, right? I think 
 that's the problem I ran into with my attempt. With your code, 
 I get an error about void:

   string []x = split(file.readln.idup, " ");
   x.each((ref s) => s.each((ref n) => n.stripRight()));

 You need to put an exclamation point after 'each' to pass the 
 function as a template parameter:

     x.each!((ref s) => s.each!((ref n) => n.stripRight()));

DOH.

But now I'm here:

extra.d(2493): Error: template extra.map_t.load_map2.each!((ref 
s) => s.each!((ref n) => n.stripRight())).each cannot deduce 
function from argument types !()(string[]), candidates are:

/usr/include/dmd/phobos/std/algorithm/iteration.d(899):        
extra.map_t.load_map2.each!((ref s) => s.each!((ref n) => 
n.stripRight())).each(Range)(Range r) if 
(!isForeachIterable!Range && (isRangeIterable!Range || 
__traits(compiles, typeof(r.front).length)))

/usr/include/dmd/phobos/std/algorithm/iteration.d(934):        
extra.map_t.load_map2.each!((ref s) => s.each!((ref n) => 
n.stripRight())).each(Iterable)(auto ref Iterable r) if 
(isForeachIterable!Iterable || __traits(compiles, 
Parameters!(Parameters!(r.opApply))))


From:

string []x = split(file.readln.idup, " ");
import std.algorithm;
x.each!((ref s) => s.each!((ref n) => n.stripRight()));

Looks like it can't tell if it's a Range or... an auto ref 
Iterable?

Apr 12 2018

=?UTF-8?Q?Ali_=c3=87ehreli?= <acehreli yahoo.com> writes:

On 04/12/2018 02:22 PM, Chris Katko wrote:
 On Thursday, 12 April 2018 at 21:17:30 UTC, Paul Backus wrote:
 On Thursday, 12 April 2018 at 20:34:40 UTC, Chris Katko wrote:
 But each doesn't return anything, it mutates, right? I think that's 
 the problem I ran into with my attempt. With your code, I get an 
 error about void:

   string []x = split(file.readln.idup, " ");
   x.each((ref s) => s.each((ref n) => n.stripRight()));

 You need to put an exclamation point after 'each' to pass the function 
 as a template parameter:

     x.each!((ref s) => s.each!((ref n) => n.stripRight()));

 
 DOH.
 
 But now I'm here:
 
 extra.d(2493): Error: template extra.map_t.load_map2.each!((ref s) => 
 s.each!((ref n) => n.stripRight())).each cannot deduce function from 
 argument types !()(string[]), candidates are:
 
 /usr/include/dmd/phobos/std/algorithm/iteration.d(899): 
 extra.map_t.load_map2.each!((ref s) => s.each!((ref n) => 
 n.stripRight())).each(Range)(Range r) if (!isForeachIterable!Range && 
 (isRangeIterable!Range || __traits(compiles, typeof(r.front).length)))
 
 /usr/include/dmd/phobos/std/algorithm/iteration.d(934): 
 extra.map_t.load_map2.each!((ref s) => s.each!((ref n) => 
 n.stripRight())).each(Iterable)(auto ref Iterable r) if 
 (isForeachIterable!Iterable || __traits(compiles, 
 Parameters!(Parameters!(r.opApply))))
 
 
 From:
 
 string []x = split(file.readln.idup, " ");
 import std.algorithm;
 x.each!((ref s) => s.each!((ref n) => n.stripRight()));
 
 Looks like it can't tell if it's a Range or... an auto ref Iterable?

Assuming the file "2darray" includes

0 1 15 0 0
2 12 1 0 0
0 1 0 10 0

the following program produces a range of ranges that produce the 
intended integer values:

import std.stdio;
import std.algorithm;
import std.conv;

void main() {
     auto file = File("2darray");
     auto r = file.byLine.map!(l => l.splitter.map!(to!int));
     writefln("%(%s\n%)", r);
}

The output is

[0, 1, 15, 0, 0]
[2, 12, 1, 0, 0]
[0, 1, 0, 10, 0]

If you want to produce an actual 2d array, you can add two .array calls 
at that chain:

     import std.range : array;
     auto r = file.byLine.map!(l => l.splitter.map!(to!int).array).array;

Ali

Apr 12 2018

Seb <seb wilzba.ch> writes:

On Thursday, 12 April 2018 at 15:38:34 UTC, Chris Katko wrote:
 I googled but couldn't find any clear solution.

 I've got a 2-D array of strings read from a text file I parsed. 
 So it's like

 0 1 15 0 0
 2 12 1 0 0
 ...
 0 1 0 10 0

 They come in with spaces, so I join into an array between them. 
 But then the last ones have a newline \n on the end, which 
 explodes the to! type conversion.

 If it was a single array, I could simply do:

 string [25] test;

   test.each((ref n) => n.stripRight());

 But how do you do that when it's a 2-D array?

 I'm looking not just for this case, but the generate case of 
 iterating / applying filters to 2-D arrays.

 Thanks.

Use mir for multi-dimensional arrays: 
https://github.com/libmir/mir-algorithm

It used to be part of phobos, but was removed as phobos's release 
cycle is too stable and dub gives much more flexibility.

Apr 12 2018

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digitalmars.D.learn - Using iteration / method chaining / etc on multi-dimensional arrays