I want to be able to construct (+; (+; `a; `b); `c) given a list of `a`b`c
Similarly if I have a list of `a`b`c`d, I want to be able to construct another nest and so on and so fourth.
I've been trying to use scan but I cant get it right
q)fsum:(+;;)/
enlist[+;;]/
q)fsum `a`b`c`d
+
(+;(+;`a;`b);`c)
`d
If you only want the raw parse tree output, one way is to form the equivalent string and use parse. This isn't recommended for more complex examples, but in this case it is clear.
{parse "+" sv string x}[`a`b`c`d]
+
`d
(+;`c;(+;`b;`a))
If you are looking to use this in a functional select, we can use +/ instead of adding each column individually, like how you specified in your example
q)parse"+/[(a;b;c;d)]"
(/;+)
(enlist;`a;`b;`c;`d)
q)f:{[t;c] ?[t;();0b;enlist[`res]!enlist (+/;(enlist,c))]};
q)t:([]a:1 2 3;b:4 5 6;c:7 8 9;d:10 11 12)
q)f[t;`a`b`c]
res
---
12
15
18
q)f[t;`a`b]
res
---
5
7
9
q)f[t;`a`b`c]~?[t;();0b;enlist[`res]!enlist (+;(+;`a;`b);`c)]
1b
You can also get the sum by indexing directly to return a list of each column values and sum over these. We use (), to turn any input into a list, otherwise it will sum the values in that single column and return only a single value
q)f:{[t;c] sum t (),c}
q)f[t;`a`b`c]
12 15 18
I currently have a query that gives me the count of each label:
g.V().group().by(label).by(count())
However this is resulting in a column for each label. I want to project two columns "Entity Type" and "Count" and count the number of each label. So far, this is all I have but it is incorrect:
g.V().project('Entity Type','Count')
.by(label)
.by(groupCount())
First, group().by(label).by(count) can be simplified to groupCount().by(label).
To reshape the result you only need a simple projection:
g.V().
groupCount().
by(label).
unfold().
project('Entity Type','Count').
by(keys).
by(values)
I currently have a 4x3500 cell array. First row is a single number, 2 row is a single string, 3rd and 4th rows are also single numbers.
Ex:
1 1 2 3 3 4 5 5 5 6
hi no ya he ........ % you get the idea
28 34 18 0 3 ......
55 2 4 42 24 .....
I would like to be able to select all columns that have a certain value in the first row. ie if I wanted '1' as the first row value, it would return
1 1
hi no
28 34
55 2
Then I would like to sort based on the 2nd row's string. ie if I wanted to have'hi', it would return:
1
hi
28
55
I have attempted to do:
variable = cellArray{:,find(cellArray{1,:} == 1)}
However I keep getting:
Error using find
Too many input arguments.
or
Error using ==
Too many input arguments.
Any help would be much appreciated! :)
{} indexing will return a comma separated list which will provide multiple outputs. When you pass this to find, it's the same as passing each element of your cell array as a separate input. This is what leads to the error about to many input arguments.
You will want to surround the comma-separated list with [] to create an array or numbers. Also, you don't need find because you can just use logical indexing to grab the columns you want. Additionally, you will want to index using () to grab the relevant rows, again to avoid the comma-separated list.
variable = cellArray(:, [cellArray{1,:}] == 1)
I know this is elementary but I can't seem to figure it out, even after reading other posts.
In a dataset, I want to convert an entire column into a date. The current class is factor.
The value in the field looks like this 12/25/2012
This is what I've tried.
C$DateofDeath=as.Date(C$DateofDeath,'%m/%d/%Y')
Error in as.Date.default(C$DateofDeath, "%m/%d/%Y") :
do not know how to convert 'C$DateofDeath' to class “Date”
C$DateofDeath=as.Date(C$DateofDeath,"%m/%d/%Y")
Error in as.Date.default(C$DateofDeath, "%m/%d/%Y") :
do not know how to convert 'C$DateofDeath' to class “Date”
Claims$DateofDeath=strptime(as.character(Claims$DateofDeath),format= '%m/%d/%Y')
Error in `$<-.data.frame`(`*tmp*`, "DateofDeath", value = list(sec = numeric(0), :
replacement has 0 rows, data has 71616
Claims$DateofDeath=strptime(as.character(Claims$DateofDeath),format= "%m/%d/%Y")
Error in `$<-.data.frame`(`*tmp*`, "DateofDeath", value = list(sec = numeric(0), :
replacement has 0 rows, data has 71616
Use as.POSIXct
C$DateOfDeath<-as.POSIXct(as.character(C$DateOfDeath), format = "%d/%m/%Y")
There are lots of R experts here but you have to specify R as one of your tags to get them to notice your question.
Looks like you have tried a bunch of combinations but not the right one.
> C <- data.frame(DateofDeath="12/25/2012",other=TRUE)
> as.Date(as.character(C$DateofDeath),format="%m/%d/%Y")
[1] "2012-12-25"
Notice that as.Date() takes a character input, not a factor. So you need to convert to character, then to Date.
Your strptime() versions seem fine to me except that you call are referring to the dataframe Claims instead of C. Actually strptime() should convert the factor to character for you, so you don't need the as.character() part with those.
I have the following dataset containing information about countries
5,1,648,16,10,2,0,3,5,1,1,0,1,1,1,0,0,0,0,0,1,0,0,1,0,0,
3,1,29,3,6,6,0,0,3,1,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,
4,1,2388,20,8,2,2,0,3,1,1,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,
...
The sixth column in each row indicates the main religion of the country: 0 is catholic, 1 is other christian, 2 is muslim, etc. Some of the other data is about if different colors are present in the flag of the country symbols they contain, and so on.
The description of the data can be found here. I have removed the string data columns though so it doesn't fit exactly like the information shown.
My problem is that I want to use co-variance matrices and Pearson correlation to see if, for example, the fact that a flag has the color red in it will tell anything about if the religion of that country has a bigger chance of being something than something else. But since the religion is enumerated, I am a bit lost on how to progress with this problem.
Your problem is that, despite the fact that your data is ordered, this order is arbitrary. The "distance" between "muslim" (enum val=1) to "hindu" (enum val=3) is not 2.
The most straight-forward way of tackling this issue is to convert enum values to binary indicator vectors:
Suppose you have
enum {
Catholic = 0
Protestant,
Muslim,
Jewish,
Hindu,
...
NumOfRel };
You replace the single entry of enum val with a binary vector of length NumOfRel with zeros everywhere except for a single 1 at the appropriate place:
For a Protestant entry, you'll have the following binary vector:
[ 0 1 0 0 ... ]
For a Jewish:
[ 0 0 0 1 0 ... ]
And so on...
This way, the "distance" between different religions is always 1.