I work on a project with MaxMSP where I have multiple colls. I want to combine all the lists in there in one single coll. Is there a way to do that directly without unpacking and repacking everything?
In order to be more clear, let’s say I have two colls, with the first one being:
0, 2
1, 4
2, 4
….
99, 9
while the second one is:
100, 8
101, 4
…
199, 7
I would like the final coll to be one list from 0-199.
Please keep in mind I don’t want to unpack everything ( with uzi for instance) cause my lists are very long and I find that it is problematic for the cpu to use colls with such long lists.That’s why I broke my huge list into sublists/subcolls in the first place
Hope that’s clear enough.
If the two colls do not have overlapping indices, then you can just dump one into the other, like this:
----------begin_max5_patcher----------
524.3ocyU0tSiCCD72IOEQV7ybnZmFJ28pfPUNI6AlKwIxeTZEh28ydsCDNB
hzdGbTolTOd20yXOd6CoIjp98flj8irqxRRdHMIAg7.IwwIjN995VtFCizAZ
M+FfjGly.6MHdisaXDTZ6DxVvfYvhfCbS8sB4MaUPsIrhWxNeUdFsf5esFex
bPYW+bc5slwBQinhFbA6qt6aaFWwPXlCCPnxDxSEQaNzhnDhG3wzT+i7+R4p
AS1YziUvTV44W3+r1ozxUnrKNdYW9gKaIbuagdkpGTv.HalU1z26bl8cTpkk
GufK9eI35911LMT2ephtnbs+0l2ybu90hl81hNex241.hHd1usga3QgGUteB
qDoYQdDYLpqv3dJR2L+BNLQodjc7VajJzrqivgs5YSkMaprkjZwroVLI03Oc
0HtKv2AMac6etChsbiQIprlPKto6.PWEfa0zX5+i8L+TnzlS7dBEaLPC8GNN
OC8qkm4MLMKx0Pm21PWjugNuwg9A6bv8URqP9m+mJdX6weocR2aU0imPwyO+
cpHiZ.sQH4FQubRLtt+YOaItUzz.3zqFyRn4UsANtZVa8RYyKWo4YSwmFane
oXSwBXC6SiMaV.anmHaBlZ9vvNPoikDIhqa3c8J+vM43PgLLDqHQA6Diwisp
Hbkqimwc8xpBMc1e4EjPp8MfRZEw6UtU9wzeCz5RFED
-----------end_max5_patcher-----------
mzed's answer works, as stated if the lists have no overlapping indices which they shouldn't based on the design you specify.
If you are treating your 'huge list' as multiple lists, or vice versa, that might help come up with an answer. One question some may ask is "why are you merging it again?"
you consider your program to have one large list
that large list is really an interface that handles how you interact with several sub-lists for efficiency sake
the interface to your data persistence (the lists) for storing and retrieval then acts like one large list but works with several under-the-hood
an insertion and retrieval mechanism for handling the multiple lists as one list should exist for your interface then
save and reload the sublists individually as well
If you wrap this into a poly~, the voice acts as the sublist, so when I say voice I basically mean sublist:
You could use a universal send/receive in and out of a poly~ abstraction that contains your sublist's unique coll, the voice# from poly~ can append uniquely to your sublist filename that is reading/saving to for that voice's [coll].
With that set up, you could specify the number of sublists (voices) and master list length you want in the poly~ arguments like:
[poly~ sublist_manager.maxpat 10 1000] // 10 sublists emulating a 1000-length list
The math for index lookup is:
//main variables for master list creation/usage
master_list_length = 1000
sublist_count = 10
sublist_length = master_list_length/sublist_count;
//variables created when inserting/looking up an index
sublist_number = (desired_index/sublist_count); //integer divide to get the base sublist you'll be performing the lookup in
sublist_index = (desired_index%sublist_length); //actual index within your sublist to access
If the above ^ is closer to what you're looking for I can work on a patch for that. cheers
Related
I tried using cmp(list1, list2) to learn it's no longer supported in Python 3.3. I've tried many other complex approaches, but none have worked.
I have two lists of which both contain just words and I want it to check to see how many words feature in both and return the number for how many.
You can find the length of the set intersection using & like this:
len(set(list1) & set(list2))
Example:
>>>len(set(['cat','dog','pup']) & set(['rat','cat','wolf']))
1
>>>set(['cat','dog','pup']) & set(['rat','cat','wolf'])
{'cat'}
Alternatively, if you don't want to use sets for some reason, you can always use collections.Counter, which supports most multiset operations:
>>> from collections import Counter
>>> print(list((Counter(['cat','dog','wolf']) & Counter(['pig','fish','cat'])).elements()))
['cat']
If you just want the count of how many words are common
common = sum(1 for i in list1 if i in list2)
If you actually want to get a list of the shared words
common_words = set(list1).intersection(list2)
I have two counts, calculated as follows:
1)g.V().hasLabel('brand').where(__.inE('client_brand').count().is(gt(0))).count()
2)g.V().hasLabel('brand').count()
and I want to get one line of code that results in the first count divided by the second.
Here's one way to do it:
g.V().hasLabel('brand').
fold().as('a','b').
math('a/b').
by(unfold().where(inE('client_brand')).count())
by(unfold().count())
Note that I simplify the first traversal to just .where(inE('client_brand')).count() since you only care to count that there is at least one edge, there's no need to count them all and do a compare.
You could also union() like:
g.V().hasLabel('brand').
union(where(inE('client_brand')).count(),
count())
fold().as('a','b').
math('a/b').
by(limit(local,1))
by(tail(local))
While the first one was a bit easier to read/follow, I guess the second is nicer because it only stores a list of the two counts whereas, the first stores a list of all the "brand" vertices which would be more memory intensive I guess.
Yet another way, provided by Daniel Kuppitz, that uses groupCount() in an interesting way:
g.V().hasLabel('brand').
groupCount().
by(choose(inE('client_brand'),
constant('a'),
constant('b'))).
math('a/(a+b)')
The following solution that uses sack() step shows why we have math() step:
g.V().hasLabel('brand').
groupCount().
by(choose(inE('client_brand'),
constant('a'),
constant('b'))).
sack(assign).
by(coalesce(select('a'), constant(0))).
sack(mult).
by(constant(1.0)). /* we need a double */
sack(div).
by(select(values).sum(local)).
sack()
If you can use lambdas then:
g.V().hasLabel('brand').
union(where(inE('client_brand')).count(),
count())
fold().
map{ it.get()[0]/it.get()[1]}
This is what worked for me:
g.V().limit(1).project('client_brand_count','total_brands')
.by(g.V().hasLabel('brand')
.where(__.inE('client_brand').count().is(gt(0))).count())
.by(g.V().hasLabel('brand').count())
.map{it.get().values()[0] / it.get().values()[1]}
.project('brand_client_pct')
I'm just starting to learn Perl and I have to do an exercise containing references.
I have to create a program, that constructs a list with two sided references, that are are received as command line arguments. At the beginning of the program, there is only one element in the list - 0. To go through the list, reference is being used, that references to the only element of the list at the moment - 0. The arguments of the command line are being read one by one and added right behind the element, that is being referenced to. When one argument is added, the reference slides one element to the right(it references to the newly added element). There are also two special elements - + and -. + allows the reference to move one element to the right, and - one element to the left. Also, it is important that the reference would not go beyond the list limit.
The output is all the arguments in the correct order of the list.
Additional requirements are that the list must be created by using hashes, that contain links to neighbouring elements. Also, I cannot use arrays to store the whole list.
There are a few examples to make it easier to grasp the concept, this is the most useful one:
./3.pl A D - B C + E
0 A B C D E
All I've got now is just the start of the program, it is nowhere near done and doesn't compile, but I can't figure out where to go from there. I've tried looking for some information about two-sided references(I'm not sure if I'm translating it correctly), but I can't seem to find anything. Any information about two-sided references or any tips how to start writing this program properly would be very appreciated.
My code:
#!/usr/bin/perl
use strict;
use warnings;
my $A= {
value=>'0',
prev=>'undef',
next=>'$B'
};
my $B= {
value=>'0',
prev=>'$A',
next=>'$C'
};
my $C= {
value=>'0',
prev=>'$B',
next=>'undef'
};
for my $smbl(0..#$ARGV) {
$A-> {value} = $ARGV[$smbl];
$Α-> {next} = $ARGV[$smbl+1];
}
First of all, what you are building is called a doubly linked list.
Let me tell you the biggest trick for working with linked lists: Create a dummy "head" node and a dummy "tail" node. You won't print their values, but having them will greatly reduce the number of special cases in your code, making it so much simpler!
At the core, you will have three "pointers" (references).
$head points to the first node of the list.
$tail points to the last node of the list.
$cursor initially points to the node in $tail. New nodes will be inserted before this node.
When processing +, two different situations you need to handle:
$cursor == $tail: Error! Cursor moved beyond end of list.
$cursor != $tail: Point $cursor to the node following the one it references.
When processing -, there are two different situations you need to handle:
$cursor->{prev} == $head: Error! Cursor moved beyond start of list.
$cursor->{prev} != $head: Point $cursor to the node preceding the one it references.
When processing inserting nodes, no checks need to be performed because of the dummy nodes!
Sorry for the length, but a picture is worth 1000 words:
In ASP.NET MVC 2, the input form field "name" attribute must contain exactly the syntax below that you would use to reference the object in C# in order to bind it back to the object upon post back. That said, if you have an object like the following where it contains multiple Orders having multiple OrderLines, the names would look and work well like this (case sensitive):
This works:
Order[0].id
Order[0].orderDate
Order[0].Customer.name
Order[0].Customer.Address
Order[0].OrderLine[0].itemID // first order line
Order[0].OrderLine[0].description
Order[0].OrderLine[0].qty
Order[0].OrderLine[0].price
Order[0].OrderLine[1].itemID // second order line, same names
Order[0].OrderLine[1].description
Order[0].OrderLine[1].qty
Order[0].OrderLine[1].price
However we want to add order lines and remove order lines at the client browser. Apparently, the indexes must start at zero and contain every consecutive index number to N.
The black belt ninja Phil Haack's blog entry here explains how to remove the [0] index, have duplicate names, and let MVC auto-enumerate duplicate names with the [0] notation. However, I have failed to get this to bind back using a nested object:
This fails:
Order.id // Duplicate names should enumerate at 0 .. N
Order.orderDate
Order.Customer.name
Order.Customer.Address
Order.OrderLine.itemID // And likewise for nested properties?
Order.OrderLine.description
Order.OrderLine.qty
Order.OrderLine.price
Order.OrderLine.itemID
Order.OrderLine.description
Order.OrderLine.qty
Order.OrderLine.price
I haven't found any advice out there yet that describes how this works for binding back nested ViewModels on post. Any links to existing code examples or strict examples on the exact names necessary to do nested binding with ILists?
Steve Sanderson has code that does this sort of thing here, but we cannot seem to get this to bind back to nested objects. Anything not having the [0]..[n] AND being consecutive in numbering simply drops off of the return object.
Any ideas?
We found a work around, by using the following:
Html.EditorFor(m => m, "ViewNameToUse", "FieldPrefix")
Where FieldPrefix is the "object[0]". This is hardly ideal, but it certainly works pretty well. It's simple and elegant.
I'm making a code generation script for UN/LOCODE system and the database has unique 3 letter/number codes in every country. So for example the database contains "EE TLL", EE being the country (Estonia) and TLL the unique code inside Estonia, "AR TLL" can also exist (the country code and the 3 letter/number code are stored separately). Codes are in capital letters.
The database is fairly big and already contains a huge number of locations, the user has also the possibility of entering the 3 letter/number him/herself (which will be checked against the database before submission automatically).
Finally neither 0 or 1 may be used (possible confusion with O and I).
What I'm searching for is the most efficient way to pick the next available code when none is provided.
What I've came up with:
I'd check with AAA till 999, but then for each code it would require a new query (slow?).
I could store all the 40000 possibilities in an array and subtract all the used codes that are already in the database... but that uses too much memory IMO (not sure what I'm talking about here actually, maybe 40000 isn't such a big number).
Generate a random code and hope it doesn't exist yet and see if it does, if it does start over again. That's just risk taking.
Is there some magic MySQL query/PHP script that can get me the next available code?
I will go with number 2, it is simple and 40000 is not a big number.
To make it more efficient, you can store a number representing each 3-letter code. The conversion should be trivial because you have a total of 34 (A-Z, 2-9) letters.
I would for option 1 (i.e. do a sequential search), adding a table that gives the last assigned code per country (i.e. such that AAA..code are all assigned already). When assigning a new code through sequential scan, that table gets updated; for user-assigned codes, it remains unmodified.
If you don't want to issue repeated queries, you can also write this scan as a stored routine.
To simplify iteration, it might be better to treat the three-letter codes as numbers (as Shawn Hsiao suggests), i.e. give a meaning to A-Z = 0..25, and 2..9 = 26..33. Then, XYZ is the number X*34^2+Y*34+Z == 23*1156+24*34+25 == 27429. This should be doable using standard MySQL functions, in particular using CONV.
I went with the 2nd option. I was also able to make a script that will try to match as close as possible the country name, for example for Tartu it will try to match T** then TA* and if possible TAR, if not it will try TAT as T is the next letter after R in Tartu.
The code is quite extensive, I'll just post the part that takes the first possible code:
$allowed = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ23456789';
$length = strlen($allowed);
$codes = array();
// store all possibilities in a huge array
for($i=0;$i<$length;$i++)
for($j=0;$j<$length;$j++)
for($k=0;$k<$length;$k++)
$codes[] = substr($allowed, $i, 1).substr($allowed, $j, 1).substr($allowed, $k, 1);
$used = array();
$query = mysql_query("SELECT code FROM location WHERE country = '$country'");
while ($result = mysql_fetch_array($query))
$used[] = $result['code'];
$remaining = array_diff($codes, $used);
$code = $remaining[0];
Thanks for your opinion, this will be the key to transport codes all over the world :)