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So, I'm using R to try and do a phylogenetic PCA on a dataset that I have using the phyl.pca function from the phytools package. However, I'm having issues organising my data in a way that the function will accept! And that's not all: I did a bit of experimenting and I know that there are more issues further down the line, which I will get into...
Getting straight to the issue, here's the data frame (with dummy data) that I'm using:
>all
Taxa Tibia Feather
1 Microraptor 138 101
2 Microraptor 139 114
3 Microraptor 145 141
4 Anchiornis 160 81
5 Anchiornis 14 NA
6 Archaeopteryx 134 82
7 Archaeopteryx 136 71
8 Archaeopteryx 132 NA
9 Archaeopteryx 14 NA
10 Scansoriopterygidae 120 85
11 Scansoriopterygidae 116 NA
12 Scansoriopterygidae 123 NA
13 Sapeornis 108 NA
14 Sapeornis 112 86
15 Sapeornis 118 NA
16 Sapeornis 103 NA
17 Confuciusornis 96 NA
18 Confuciusornis 107 30
19 Confuciusornis 148 33
20 Confuciusornis 128 61
The taxa are arranged into a tree (called "tree") with Microraptor being the most basal and then progressing in order through to Confuciusornis:
>summary(tree)
Phylogenetic tree: tree
Number of tips: 6
Number of nodes: 5
Branch lengths:
mean: 1
variance: 0
distribution summary:
Min. 1st Qu. Median 3rd Qu. Max.
1 1 1 1 1
No root edge.
Tip labels: Confuciusornis
Sapeornis
Scansoriopterygidae
Archaeopteryx
Anchiornis
Microraptor
No node labels.
And the function:
>phyl.pca(tree, all, method="BM", mode="corr")
And this is the error that is coming up:
Error in phyl.pca(tree, all, method = "BM", mode = "corr") :
number of rows in Y cannot be greater than number of taxa in your tree
Y being the "all" data frame. So I have 6 taxa in my tree (matching the 6 taxa in the data frame) but there are 20 rows in my data frame. So I used this function:
> all_agg <- aggregate(all[,-1],by=list(all$Taxa),mean,na.rm=TRUE)
And got this:
Group.1 Tibia Feather
1 Anchiornis 153 81
2 Archaeopteryx 136 77
3 Confuciusornis 120 41
4 Microraptor 141 119
5 Sapeornis 110 86
6 Scansoriopterygidae 120 85
It's a bit odd that the order of the taxa has changed... Is this ok?
In any case, I converted it into a matrix:
> all_agg_matrix <- as.matrix(all_agg)
> all_agg_matrix
Group.1 Tibia Feather
[1,] "Anchiornis" "153" "81"
[2,] "Archaeopteryx" "136" "77"
[3,] "Confuciusornis" "120" "41"
[4,] "Microraptor" "141" "119"
[5,] "Sapeornis" "110" "86"
[6,] "Scansoriopterygidae" "120" "85"
And then used the phyl.pca function:
> phyl.pca(tree, all_agg_matrix, method = "BM", mode = "corr")
[1] "Y has no names. function will assume that the row order of Y matches tree$tip.label"
Error in invC %*% X : requires numeric/complex matrix/vector arguments
So, now the order that the function is considering taxa in is all wrong (but I can fix that relatively easily). The issue is that phyl.pca doesn't seem to believe that my matrix is actually a matrix. Any ideas why?
I think you may have bigger problems. Most phylogenetic methods, I suspect including phyl.pca, assume that traits are fixed at the species level (i.e., they don't account for within-species variation). Thus, if you want to use phyl.pca, you probably need to collapse your data to a single value per species, e.g. via
dd_agg <- aggregate(dd[,-1],by=list(dd$Taxa),mean,na.rm=TRUE)
Extract the numeric columns and label the rows properly so that phyl.pca can match them up with the tips correctly:
dd_mat <- dd_agg[,-1]
rownames(dd_mat) <- dd_agg[,1]
Using these aggregated data, I can make up a tree (since you didn't give us one) and run phyl.pca ...
library(phytools)
tt <- rcoal(nrow(dd_agg),tip.label=dd_agg[,1])
phyl.pca(tt,dd_mat)
If you do need to do an analysis that takes within-species variation into account you might need to ask somewhere more specialized, e.g. the r-sig-phylo#r-project.org mailing list ...
The answer posted by Ben Bolker seems to work whereby the data (called "all") is collapsed into a single value per species before creating a matrix and running the function. As per so:
> all_agg <- aggregate(all[,-1],by=list(all$Taxa),mean,na.rm=TRUE)
> all_mat <- all_agg[,-1]
> rownames(all_mat) <- all_agg[,1]
> phyl.pca(tree,all_mat, method= "lambda", mode = "corr")
Thanks to everyone who contributed an answer and especially Ben! :)
I have the following data frame which I called ozone:
Ozone Solar.R Wind Temp Month Day
1 41 190 7.4 67 5 1
2 36 118 8.0 72 5 2
3 12 149 12.6 74 5 3
4 18 313 11.5 62 5 4
5 NA NA 14.3 56 5 5
6 28 NA 14.9 66 5 6
7 23 299 8.6 65 5 7
8 19 99 13.8 59 5 8
9 8 19 20.1 61 5 9
I would like to extract the highest value from ozone, Solar.R, Wind...
Also, if possible how would I sort Solar.R or any column of this data frame in descending order
I tried
max(ozone, na.rm=T)
which gives me the highest value in the dataset.
I have also tried
max(subset(ozone,Ozone))
but got "subset" must be logical."
I can set an object to hold the subset of each column, by the following commands
ozone <- subset(ozone, Ozone >0)
max(ozone,na.rm=T)
but it gives the same value of 334, which is the max value of the data frame, not the column.
Any help would be great, thanks.
Similar to colMeans, colSums, etc, you could write a column maximum function, colMax, and a column sort function, colSort.
colMax <- function(data) sapply(data, max, na.rm = TRUE)
colSort <- function(data, ...) sapply(data, sort, ...)
I use ... in the second function in hopes of sparking your intrigue.
Get your data:
dat <- read.table(h=T, text = "Ozone Solar.R Wind Temp Month Day
1 41 190 7.4 67 5 1
2 36 118 8.0 72 5 2
3 12 149 12.6 74 5 3
4 18 313 11.5 62 5 4
5 NA NA 14.3 56 5 5
6 28 NA 14.9 66 5 6
7 23 299 8.6 65 5 7
8 19 99 13.8 59 5 8
9 8 19 20.1 61 5 9")
Use colMax function on sample data:
colMax(dat)
# Ozone Solar.R Wind Temp Month Day
# 41.0 313.0 20.1 74.0 5.0 9.0
To do the sorting on a single column,
sort(dat$Solar.R, decreasing = TRUE)
# [1] 313 299 190 149 118 99 19
and over all columns use our colSort function,
colSort(dat, decreasing = TRUE) ## compare with '...' above
To get the max of any column you want something like:
max(ozone$Ozone, na.rm = TRUE)
To get the max of all columns, you want:
apply(ozone, 2, function(x) max(x, na.rm = TRUE))
And to sort:
ozone[order(ozone$Solar.R),]
Or to sort the other direction:
ozone[rev(order(ozone$Solar.R)),]
Here's a dplyr solution:
library(dplyr)
# find max for each column
summarise_each(ozone, funs(max(., na.rm=TRUE)))
# sort by Solar.R, descending
arrange(ozone, desc(Solar.R))
UPDATE: summarise_each() has been deprecated in favour of a more featureful family of functions: mutate_all(), mutate_at(), mutate_if(), summarise_all(), summarise_at(), summarise_if()
Here is how you could do:
# find max for each column
ozone %>%
summarise_if(is.numeric, funs(max(., na.rm=TRUE)))%>%
arrange(Ozone)
or
ozone %>%
summarise_at(vars(1:6), funs(max(., na.rm=TRUE)))%>%
arrange(Ozone)
In response to finding the max value for each column, you could try using the apply() function:
> apply(ozone, MARGIN = 2, function(x) max(x, na.rm=TRUE))
Ozone Solar.R Wind Temp Month Day
41.0 313.0 20.1 74.0 5.0 9.0
Another way would be to use ?pmax
do.call('pmax', c(as.data.frame(t(ozone)),na.rm=TRUE))
#[1] 41.0 313.0 20.1 74.0 5.0 9.0
There is a package matrixStats that provides some functions to do column and row summaries, see in the package vignette, but you have to convert your data.frame into a matrix.
Then you run: colMaxs(as.matrix(ozone))
max(may$Ozone, na.rm = TRUE)
Without $Ozone it will filter in the whole data frame, this can be learned in the swirl library.
I'm studying this course on Coursera too ~
Assuming that your data in data.frame called maxinozone, you can do this
max(maxinozone[1, ], na.rm = TRUE)
max(ozone$Ozone, na.rm = TRUE) should do the trick. Remember to include the na.rm = TRUE or else R will return NA.
Try this solution:
Oz<-subset(data, data$Month==5,select=Ozone) # select ozone value in the month of
#May (i.e. Month = 5)
summary(T) #gives caracteristics of table( contains 1 column of Ozone) including max, min ...
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!
I try to apply a function over all rows and columns of two dataframes but I don't know how to solve it with apply.
I think the following script explains what I intend to do and the way i tried to solve it. Any advice would be warmly appreciated! Please note, that the simplefunction is only intended to be an example function to keep it simple.
# some data and a function
df1<-data.frame(name=c("aa","bb","cc","dd","ee"),a=sample(1:50,5),b=sample(1:50,5),c=sample(1:50,5))
df2<-data.frame(name=c("aa","bb","cc","dd","ee"),a=sample(1:50,5),b=sample(1:50,5),c=sample(1:50,5))
simplefunction<-function(a,b){a+b}
# apply on a single row
simplefunction(df1[1,2],df2[1,2])
# apply over all colums
apply(?)
## apply over all columns and rows
# create df to receive results
df3<-df2
# loop it
for (i in 2:5)df3[i]<-apply(?)
My first mapply answer!! For your simple example you have...
mapply( FUN = `+` , df1[,-1] , df2[,-1] )
# a b c
# [1,] 60 35 75
# [2,] 57 39 92
# [3,] 72 71 48
# [4,] 31 19 85
# [5,] 47 66 58
You can extend it like so...
mapply( FUN = function(x,y,z,etc){ simplefunctioncodehere} , df1[,-1] , df2[,-1] , ... other dataframes here )
The dataframes will be passed in order to the function, so in this example df1 would be x, df2 would be y and z and etc would be some other dataframes that you specify in that order. Hopefully that makes sense. mapply will take the first row, first column values of all dataframes and apply the function, then the first row, second column of all data frames and apply the function and so on.
You can also use Reduce:
set.seed(45) # for reproducibility
Reduce(function(x,y) { x + y}, list(df1[, -1], df2[,-1]))
# a b c
# 1 53 22 23
# 2 64 28 91
# 3 19 56 51
# 4 38 41 53
# 5 28 42 30
You can just do :
df1[,-1] + df2[,-1]
Which gives :
a b c
1 52 24 37
2 65 63 62
3 31 90 89
4 90 35 33
5 51 33 45
I don't understand why I can't find a solution for this, since I feel that this is a pretty basic question. Need to ask for help, then. I want to rearrange airquality dataset by month with maximum temp value for each month. In addition I want to find the corresponding day for each monthly maximum temperature. What is the laziest (code-wise) way to do this?
I have tried following without a success:
require(reshape2)
names(airquality) <- tolower(names(airquality))
mm <- melt(airquality, id.vars = c("month", "day"), meas = c("temp"))
dcast(mm, month + day ~ variable, max)
aggregate(formula = temp ~ month + day, data = airquality, FUN = max)
I am after something like this:
month day temp
5 7 89
...
There was quite a discussion a while back about whether being lazy is good or not. Anwyay, this is short and natural to write and read (and is fast for large data so you don't need to change or optimize it later) :
require(data.table)
DT=as.data.table(airquality)
DT[,.SD[which.max(Temp)],by=Month]
Month Ozone Solar.R Wind Temp Day
[1,] 5 45 252 14.9 81 29
[2,] 6 NA 259 10.9 93 11
[3,] 7 97 267 6.3 92 8
[4,] 8 76 203 9.7 97 28
[5,] 9 73 183 2.8 93 3
.SD is the subset of the data for each group, and you just want the row from it with the largest Temp, iiuc. If you need the row number then that can be added.
Or to get all the rows where the max is tied :
DT[,.SD[Temp==max(Temp)],by=Month]
Month Ozone Solar.R Wind Temp Day
[1,] 5 45 252 14.9 81 29
[2,] 6 NA 259 10.9 93 11
[3,] 7 97 267 6.3 92 8
[4,] 7 97 272 5.7 92 9
[5,] 8 76 203 9.7 97 28
[6,] 9 73 183 2.8 93 3
[7,] 9 91 189 4.6 93 4
Another approach with plyr
require(reshape2)
names(airquality) <- tolower(names(airquality))
mm <- melt(airquality, id.vars = c("month", "day"), meas = c("temp"), value.name = 'temp')
library(plyr)
ddply(mm, .(month), subset, subset = temp == max(temp), select = -variable)
Gives
month day temp
1 5 29 81
2 6 11 93
3 7 8 92
4 7 9 92
5 8 28 97
6 9 3 93
7 9 4 93
Or, even simpler
require(reshape2)
require(plyr)
names(airquality) <- tolower(names(airquality))
ddply(airquality, .(month), subset,
subset = temp == max(temp), select = c(month, day, temp) )
how about with plyr?
max.func <- function(df) {
max.temp <- max(df$temp)
return(data.frame(day = df$Day[df$Temp==max.temp],
temp = max.temp))
}
ddply(airquality, .(Month), max.func)
As you can see, the max temperature for the month happens on more than one day. If you want different behavior, the function is easy enough to adjust.
Or if you want to use the data.table package (for instance, if speed is an issue and the data set is large or if you prefer the syntax):
library(data.table)
DT <- data.table(airquality)
DT[, list(maxTemp=max(Temp), dayMaxTemp=.SD[max(Temp)==Temp, Day]), by="Month"]
If you want to know what the .SD stands for, have a look here: SO