I am trying to create a limit order book and in one of the functions I want to return a list that sums the column 'size' for the ask dataframe and the bid dataframe in the limit order book.
The output should be...
$ask
oid price size
8 a 105 100
7 o 104 292
6 r 102 194
5 k 99 71
4 q 98 166
3 m 98 88
2 j 97 132
1 n 96 375
$bid
oid price size
1 b 95 100
2 l 95 29
3 p 94 87
4 s 91 102
Total volume: 318 1418
Where the input is...
oid,side,price,size
a,S,105,100
b,B,95,100
I have a function book.total_volumes <- function(book, path) { ... } that should return total volumes.
I tried to use aggregate but struggled with the fact that it is both ask and bid in the limit order book.
I appreciate any help, I am clearly a complete beginner. Only hear to learn :)
If there is anything more I can add to this question so is more clear feel free to leave a comment!
Related
I have a data set of counts from standard solutions passed through an instrument that analyses chemical concentrations (an ICPMS for those familiar). The data is over a range of different standards and for each standard I have four repeat measurements that I want to calculate the mean and variance of.
I'm importing the data from an excel spreadsheet and then, following some housekeeping such as getting dates and times in the right format, I split the the dataset up into a list identified by the name of the standard solution using Count11.sp<-split(Count11.raw, Count11.raw$Type). Count11.raw$Type then becomes the list element name and I have the four count results for each chemical element in that list element.
So far so good.
I find I can yield an average (mean, median etc) easily enough by identifying the list element specifically i.e. mean(Count11.sp$'Ca40') , or sapply(Count11$'Ca40', median), but what I'm not able to do is automate that in a loop so that I can calculate the means for each standard and drop that into a numerical matrix for further manipulation. I can extract the list element names with names() and I can even use a loop to make a vector of all the names and reference the specific list element using these in a for loop.
For instance Count11.sp[names(Count11.sp[i])]will extract the full list element no problem:
$`Post Ca45t`
Type Run Date 7Li 9Be 24Mg 43Ca 52Cr 55Mn 59Co 60Ni
77 Post Ca45t 1 2011-02-08 00:13:08 114 26101 4191 453525 2632 520 714 2270
78 Post Ca45t 2 2011-02-08 00:13:24 114 26045 4179 454299 2822 524 704 2444
79 Post Ca45t 3 2011-02-08 00:13:41 96 26372 3961 456293 2898 520 762 2244
80 Post Ca45t 4 2011-02-08 00:13:58 112 26244 3799 454702 2630 510 792 2356
65Cu 66Zn 85Rb 86Sr 111Cd 115In 118Sn 137Ba 140Ce 141Pr 157Gd 185Re 208Pb
77 244 1036 56 3081 44 520625 78 166 724 10 0 388998 613
78 250 982 70 3103 46 526154 76 174 744 16 4 396496 644
79 246 1014 36 3183 56 524195 60 198 744 2 0 396024 612
80 270 932 60 3137 44 523366 70 180 824 2 4 390436 632
238U
77 24
78 20
79 14
80 6
but sapply(Count11.sp[names(count11.sp[i])produces an error message: Error in median.default(X[[i]], ...) : need numeric data
while sapply(Input$Post Ca45t, median) <'Post Ca45t' being name Count11.sp[i] i=4> does exactly what I want and produces the median value (I can clean that vector up later for medians that don't make sense) e.g.
Type Run Date 7Li 9Be 24Mg
NA 2.5 1297109612.5 113.0 26172.5 4070.0
43Ca 52Cr 55Mn 59Co 60Ni 65Cu
454500.5 2727.0 520.0 738.0 2313.0 248.0
66Zn 85Rb 86Sr 111Cd 115In 118Sn
998.0 58.0 3120.0 45.0 523780.5 73.0
137Ba 140Ce 141Pr 157Gd 185Re 208Pb
177.0 744.0 6.0 2.0 393230.0 622.5
238U
17.0
Can anyone give me any insight into how I can automate (i.e. loop through) these names to produce one median vector per list element? I'm sure there's just some simple disconnect in my logic here that may be easily solved.
Update: I've solved the problem. The way to do so is to use tapply on the original dataset with out the need to split it. tapply allows functions to be applied to data based on a user defined grouping criteria. In my case I could group according to the Count11.raw$Type and then take the mean of the data subset. tapply(Count11.raw$Type, Count11.raw[,3:ncol(Count11.raw)], mean), job done.
I have a file with data on the delivery of products to the store.I need to calculate the total number of products in the store. I want to use the knowledge of cycles to calculate the total quantity of the product in the store, but my cycle only counts the total quantity of the last product. Why?
Here is the delivery data:
"Day" "Cott.cheese, pcs." "Kefir, pcs." "Sour cream, pcs."
1 104 117 119
2 94 114 114
3 105 107 117
4 99 112 120
5 86 104 111
6 88 110 126
7 95 106 129
I put this table in the in1 variable
Here is code:
s<-0
for (p in (2:ncol(in1))){
s<-sum(in1[,p]) }
s
Not sure I understand correctly your question but if you only want to add all values of your data.frame except for the first column (Day), you just need to do this:
sum(in1[,-1])
You are rewriting the s variable each iteration, that's why it only shows the result for the last column. Try
s<-c()
for (p in 2:ncol(in1)) {
s<-c(s,sum(in1[,p]))
}
alternatively
colSums(in1[,-1])
I'm struggeling to get a good performing script for this problem: I have a table with a score, x, y. I want to sort the table by score and than build groups based on the x value. Each group should have an equal sum (not counts) of x. x is a metric number in the dataset and resembles the historic turnover of a customer.
score x y
0.436024136 3 435
0.282303336 46 56
0.532358015 24 34
0.644236597 0 2
0.99623626 0 4
0.557673456 56 46
0.08898779 0 7
0.702941303 453 2
0.415717835 23 1
0.017497461 234 3
0.426239166 23 59
0.638896238 234 86
0.629610596 26 68
0.073107526 0 35
0.85741877 0 977
0.468612039 0 324
0.740704267 23 56
0.720147257 0 68
0.965212467 23 0
a good way to do so is adding a group variable to the data.frame with cumsum! Now you can easily sum the groups with e. g. subset.
data.frame$group <-cumsum(as.numeric(data.frame$x)) %/% (ceiling(sum(data.frame$x) / 3)) + 1
remarks:
in big data.frames cumsum(as.numeric()) works reliably
%/% is a division where you get an integer back
the '+1' just let your groups start with 1 instead of 0
thank you #Ronak Shah!
I have written a function that will compare the similarity of IP addresses, and will let the user select the level of detail in the octet. for example, in the address 255.255.255.0 and 255.255.255.1, a user could specify that they only want to compare the first, first and second, first second third etc. octets.
the function is below:
did.change.ip=function(vec, detail){
counter=2
result.vec=FALSE
r.list=strsplit(vec, '.', fixed=TRUE)
for(i in vec){
if(counter>length(vec)){
break
}
first=as.numeric(r.list[[counter-1]][1:detail])
second=as.numeric(r.list[[counter]][1:detail])
if(sum(first==second)==detail){
result.vec=append(result.vec,FALSE)
}
else{
result.vec=append(result.vec,TRUE)
}
counter=counter+1
}
return(result.vec)
}
and it's really slow once the data starts getting larger. for a dataset of 500,000 rows, the system.time() results are:
user system elapsed
208.36 0.59 209.84
are there any R power users who have insight on how to write this more efficiently? I know lapply() is the preferred method for looping over vectors/dataframes, but I'm stumped as to how to access the previous element in a vector for this purpose. I've tried to sketch something out quickly, but It returns a syntax error:
test=function(vec, detail){
rlist=strsplit(vec, '.', fixed=TRUE)
r.value=vapply(rlist, function(x,detail) ifelse(x[1:detail]==x[1:detail] TRUE, FALSE))
}
I've created some sample data for testing purposes below:
stack.data=structure(list(V1 = c("247.116.209.66", "195.121.47.105", "182.136.49.12",
"237.123.100.50", "120.30.174.18", "29.85.72.70", "18.186.76.177",
"33.248.142.26", "109.97.92.50", "217.138.155.145", "20.203.156.2",
"71.1.51.190", "31.225.208.60", "55.25.129.73", "211.204.249.244",
"198.137.15.53", "234.106.102.196", "244.3.87.9", "205.242.10.22",
"243.61.212.19", "32.165.79.86", "190.207.159.147", "157.153.136.100",
"36.151.152.15", "2.254.210.246", "3.42.1.208", "30.11.229.18",
"72.187.36.103", "98.114.189.34", "67.93.180.224")), .Names = "V1", class = "data.frame", row.names = c(NA,
-30L))
Here's another solution just using base R.
did.change.ip <- function(vec, detail=4){
ipv <- scan(text=paste(vec, collapse="\n"),
what=c(replicate(detail, integer()), replicate(4-detail,NULL)),
sep=".", quiet=TRUE)
c(FALSE, rowSums(vapply(ipv[!sapply(ipv, is.null)],
diff, integer(length(vec)-1))!=0)>0)
}
Here we use scan() to break up the ip address into numbers. Then we we look down each octet for differences using diff. It seems this is faster than the original proposal, but slightly slower than #josilber's stringr solution (using microbenchmark with 3,000 ip addresses)
Unit: milliseconds
expr min lq median uq max neval
orig 35.251886 35.716921 36.019354 36.700550 90.159992 100
scan 2.062189 2.116391 2.170110 2.236658 3.563771 100
strngr 2.027232 2.075018 2.136114 2.200096 3.535227 100
The simplest way I can think of to do this is to build a transformed vector that only includes the parts of the IP you want. Then it's a one-liner to check if each element is equal to the one before it:
library(stringr)
did.change.josilber <- function(vec, detail) {
s <- str_extract(vec, paste0("^(\\d+\\.){", detail, "}"))
return(s != c(s[1], s[1:(length(s)-1)]))
}
This seems reasonably efficient for 500,000 rows:
set.seed(144)
big.vec <- sample(stack.data[,1], 500000, replace=T)
system.time(did.change.josilber(big.vec, 3))
# user system elapsed
# 0.527 0.030 0.554
The biggest issue with your code is that you call append each iteration, which requires reallocation of your vector 500,000 times. You can read more about this in the second circle of the R inferno.
Not sure if all you want is counts, but this is potentially a solution:
library(dplyr)
library(tidyr)
# split ip addresses into "octets"
octets <- stack.data %>%
separate(V1,c("first","second","third","fourth"))
# how many shared both their first and second octets?
octets %>%
group_by(first,second) %>%
summarize(n = n())
first second n
1 109 97 1
2 120 30 1
3 157 153 1
4 18 186 1
5 182 136 1
6 190 207 1
7 195 121 1
8 198 137 1
9 2 254 1
10 20 203 1
11 205 242 1
12 211 204 1
13 217 138 1
14 234 106 1
15 237 123 1
16 243 61 1
17 244 3 1
18 247 116 1
19 29 85 1
20 3 42 1
21 30 11 1
22 31 225 1
23 32 165 1
24 33 248 1
25 36 151 1
26 55 25 1
27 67 93 1
28 71 1 1
29 72 187 1
30 98 114 1
edited to improve the quality of the question as a result of the (wholly appropriate) spanking received by Spacedman!
I have a k-nearest neighbors object (an igraph) which I created as such, by using the file I have uploaded here:
I performed the following operations on the data, in order to create an adjacency matrix of distances between observations:
W <- read.csv("/path/sim_matrix.csv")
W <- W[, -c(1,3)]
W <- scale(W)
sim_matrix <- dist(W, method = "euclidean", upper=TRUE)
sim_matrix <- as.matrix(sim_matrix)
mygraph <- nng(sim_matrix, k=10)
This give me a nice list of vertices and their ten closest neighbors, a small sample follows:
1 -> 25 26 28 30 32 144 146 151 177 183 2 -> 4 8 32 33 145 146 154 156 186 199
3 -> 1 25 28 51 54 106 144 151 177 234 4 -> 7 8 89 95 97 158 160 170 186 204
5 -> 9 11 17 19 21 112 119 138 145 158 6 -> 10 12 14 18 20 22 147 148 157 194
7 -> 4 13 123 132 135 142 160 170 173 174 8 -> 4 7 89 90 95 97 158 160 186 204
So far so good.
What I'm struggling with, however, is how to to get access to the values for the weights between the vertices that I can do meaningful calculations on. Shouldn't be so hard, this is a common thing to want from graphs, no?
Looking at the documentation, I tried:
degree(mygraph)
which gives me the sum of the weights for each node. But I don't want the sum, I want the raw data, so I can do my own calculations.
I tried
get.data.frame(mygraph,"E")[1:10,]
but this has none of the distances between nodes:
from to
1 1 25
2 1 26
3 1 28
4 1 30
5 1 32
6 1 144
7 1 146
8 1 151
9 1 177
10 1 183
I have attempted to get values for the weights between vertices out of the graph object, that I can work with, but no luck.
If anyone has any ideas on how to go about approaching this, I'd be grateful. Thanks.
It's not clear from your question whether you are starting with a dataset, or with a distance matrix, e.g. nng(x=mydata,...) or nng(dx=mydistancematrix,...), so here are solutions with both.
library(cccd)
df <- mtcars[,c("mpg","hp")] # extract from mtcars dataset
# knn using dataset only
g <- nng(x=as.matrix(df),k=5) # for each car, 5 other most similar mpg and hp
V(g)$name <- rownames(df) # meaningful names for the vertices
dm <- as.matrix(dist(df)) # full distance matrix
E(g)$weight <- apply(get.edges(g,1:ecount(g)),1,function(x)dm[x[1],x[2]])
# knn using distance matrix (assumes you have dm already)
h <- nng(dx=dm,k=5)
V(h)$name <- rownames(df)
E(h)$weight <- apply(get.edges(h,1:ecount(h)),1,function(x)dm[x[1],x[2]])
# same result either way
identical(get.data.frame(g),get.data.frame(h))
# [1] TRUE
So these approaches identify the distances from each vertex to it's five nearest neighbors, and set the edge weight attribute to those values. Interestingly, plot(g) works fine, but plot(h) fails. I think this might be a bug in the plot method for cccd.
If all you want to know is the distances from each vertex to the nearest neighbors, the code below does not require package cccd.
knn <- t(apply(dm,1,function(x)sort(x)[2:6]))
rownames(knn) <- rownames(df)
Here, the matrix knn has a row for each vertex and columns specifying the distance from that vertex to it's 5 nearest neighbors. It does not tell you which neighbors those are, though.
Okay, I've found a nng function in cccd package. Is that it? If so.. then mygraph is just an igraph object and you can just do E(mygraph)$whatever to get the names of the edge attributes.
Following one of the cccd examples to create G1 here, you can get a data frame of all the edges and attributes thus:
get.data.frame(G1,"E")[1:10,]
You can get/set individual edge attributes with E(g)$whatever:
> E(G1)$weight=1:250
> E(G1)$whatever=runif(250)
> get.data.frame(G1,"E")[1:10,]
from to weight whatever
1 1 3 1 0.11861240
2 1 7 2 0.06935047
3 1 22 3 0.32040316
4 1 29 4 0.86991432
5 1 31 5 0.47728632
Is that what you are after? Any igraph package tutorial will tell you more!