I dont know how to explain it shortly. I try my best:
I have the following example data:
Data<-data.frame(A=c(1,2,3,5,8,9,10),B=c(5.3,9.2,5,8,10,9.5,4),C=c(1:7))
and a index
Ind<-data.frame(I=c(5,6,2,4,1,3,7))
The value in Ind corresponds to the C column in the Data. Now I want to start with the first Ind value, and find the corresponding row in the Data data.frame (column C). From that row I want to go up and down and find values in column A that are in a tolerance range of 1. I want to write these values into a result dataframe add a group id column and delete it in the dataframe Data (where I found them). Then I start with the next entry in the Index dataframe Ind and so an until the data.frame Data is empty. I know how to match my Ind with column C of my Data and how to write and delete and the other stuff in a for loop, but I dont know the main point, which is my question here:
when I have found my row in the Data, how can I look up fitting values of column A in the tolerance range up and below that entry to get my Group id?
what I want to get is this result:
A B C Group
1 5.3 1 2
2 9.2 2 2
3 5 3 2
5 8 4 3
8 10 5 1
9 9.5 6 1
10 4 7 4
Maybe somebody could help me with the critical point in my question or even how to solve this issue in a fast way.
Many thanks!
Generally: avoid deleting or growing a data frame row by row inside a loop. R's memory management means that every time you add or delete a row, another copy of the data frame is made. Garbage collection will eventually discard the "old" copies of the data frame, but garbage can quickly accumulate and reduce performance. Instead, add a logical column to the Data data frame, and set "extracted" rows to TRUE. So like this:
Data$extracted <- rep(FALSE,nrow(Data))
As for your problem: I get a different set of grouping numbers, but the groups are identical.
There might be a more elegant way to do this, but this will get it done.
# store results in a separate list
res <- list()
group.counter <- 1
# loop until they're all done.
for(idx in Ind$I) {
# skip this iteration if idx is NA.
if(is.na(idx)) {
next
}
# dat.rows is a logical vector which shows the rows where
# "A" meets the tolerance requirement.
# specify the tolerance here.
mytol <- 1
# the next only works for integer compare.
# also not covered: what if multiple values of C
# match idx? do we loop over each corresponding value of A,
# i.e. loop over each value of 'target'?
target <- Data$A[Data$C == idx]
# use the magic of vectorized logical compare.
dat.rows <-
( (Data$A - target) >= -mytol) &
( (Data$A - target) <= mytol) &
( ! Data$extracted)
# if dat.rows is all false, then nothing met the criteria.
# skip the rest of the loop
if( ! any(dat.rows)) {
next
}
# copy the rows to the result list.
res[[length(res) + 1]] <- data.frame(
A=Data[dat.rows,"A"],
B=Data[dat.rows,"B"],
C=Data[dat.rows,"C"],
Group=group.counter # this value will be recycled to match length of A, B, C.
)
# flag the extraction.
Data$extracted[dat.rows] <- TRUE
# increment the group counter
group.counter <- group.counter + 1
}
# now make a data.frame from the results.
# this is the last step in how we avoid
#"growing" a data.frame inside a loop.
resData <- do.call(rbind, res)
Related
I'm looking for a way to exclude a number of answers from a length function.
This is a follow on question from Getting R Frequency counts for all possible answers In sql the syntax could be
select * from someTable
where variableName not in ( 0, null )
Given
Id <- c(1,2,3,4,5)
ClassA <- c(1,NA,3,1,1)
ClassB <- c(2,1,1,3,3)
R <- c(5,5,7,NA,9)
S <- c(3,7,NA,9,5)
df <- data.frame(Id,ClassA,ClassB,R,S)
ZeroTenNAScale <- c(0:10,NA);
R.freq = setNames(nm=c('R','freq'),data.frame(table(factor(df$R,levels=ZeroTenNAScale,exclude=NULL))));
S.freq = setNames(nm=c('S','freq'),data.frame(table(factor(df$S,levels=ZeroTenNAScale,exclude=NULL))));
length(S.freq$freq[S.freq$freq!=0])
# 5
How would I change
length(S.freq$freq[S.freq$freq!=0])
to get an answer of 4 by excluding 0 and NA?
We can use colSums,
colSums(!is.na(S.freq)[S.freq$freq!=0,])[[1]]
#[1] 4
You can use sum to calculate the sum of integers. if NA's are found in your column you could be using na.rm(), however because the NA is located in a different column you first need to remove the row containing NA.
Our solution is as follows, we remove the rows containing NA by subsetting S.freq[!is.na(S.freq$S),], but we also need the second column freq:
sum(S.freq[!is.na(S.freq$S), "freq"])
# 4
You can try na.omit (to remove NAs) and subset ( to get rid off all lines in freq equal to 0):
subset(na.omit(S.freq), freq != 0)
S freq
4 3 1
6 5 1
8 7 1
10 9 1
From here, that's straightforward:
length(subset(na.omit(S.freq), freq != 0)$freq)
[1] 4
Does it solve your problem?
Just add !is.na(S.freq$S) as a second filter:
length(S.freq$freq[S.freq$freq!=0 & !is.na(S.freq$S)])
If you want to extend it with other conditions, you could make an index vector first for readability:
idx <- S.freq$freq!=0 & !is.na(S.freq$S)
length(S.freq$freq[idx])
You're looking for values with frequency > 0, that means you're looking for unique values. You get this information directly from vector S:
length(unique(df$S))
and leaving NA aside you get answer 4 by:
length(unique(df$S[!is.na(df$S)]))
Regarding your question on how to exclude a number of items based on their value:
In R this is easily done with logical vectors as you used it in you code already:
length(S.freq$freq[S.freq$freq!=0])
you can combine different conditions to one logical vector and use it for subsetting e.g.
length(S.freq$freq[S.freq$freq!=0 & !is.na(S.freq$freq)])
Part of a function I'm working on uses the following code to take a data frame and reorder its columns on the basis of the largest (absolute) value in each column.
ord <- order(abs(apply(dfm,2,function(x) x[which(abs(x) == max(abs(x)), arr.ind = TRUE)])))
For the most part, this works fine, but with the dataset I'm working on, I occasionally get data that looks like this:
a <- rnorm(10,5,7); b <- rnorm(10,0,1); c <- rep(1,10)
dfm <- data.frame(A = a, B = b, C = c)
> dfm
A B C
1 0.6438373 -1.0487023 1
2 10.6882204 0.7665011 1
3 -16.9203506 -2.5047946 1
4 11.7160291 -0.1932127 1
5 13.0839793 0.2714989 1
6 11.4904625 0.5926858 1
7 -5.9559206 0.1195593 1
8 4.6305924 -0.2002087 1
9 -2.2235623 -0.2292297 1
10 8.4390810 1.1989515 1
When that happens, the above code returns a "non-numeric argument to mathematical function" error at the abs() step. (And if I get rid of the abs() step because I know, due to transformation, my data will be all positive, order() returns: "unimplemented type 'list' in 'orderVector1'".) This is because which() returns all the 1's in column C, which in turn makes apply() spit out a list, rather than a nice tidy vector.
My question is this: How can I make which() JUST return one value for column C in this case? Alternately, is there a better way to write this code to do what I want it to (reorder the columns of a matrix based on the largest value in each column, whether or not that largest value is duplicated) that won't have this problem?
If you want to select just the first element of the result, you can subset it with [1]:
ord <- order(abs(apply(dfm,2,function(x) x[which(abs(x) == max(abs(x)), arr.ind = TRUE)][1])))
To order the columns by their maximum element (in absolute value), you can do
dfm[order(apply(abs(dfm),2,max))]
Your code, with #CarlosCinelli's correction, should work fine, though.
I would like to select specific elements of a data.list after processing it.
To get process parameters I describe the my problem in the reproducible example.
In the example code below, I have three sets of data.list each have 5 column.
Each data.list repeat theirselves three times each and each data.list assignet to unique number called set_nbr which defines these datasets.
#to create reproducible data (this part creates three sets of data each one repeats 3 times of those of Mx, My and Mz values along with set_nbr)
set.seed(1)
data.list <- lapply(1:3, function(x) {
nrep <- 3
time <- rep(seq(90,54000,length.out=600),times=nrep)
Mx <- c(replicate(nrep,sort(runif(600,-0.014,0.012),decreasing=TRUE)))
My <- c(replicate(nrep,sort(runif(600,-0.02,0.02),decreasing=TRUE)))
Mz <- c(replicate(nrep,sort(runif(600,-1,1),decreasing=TRUE)))
df <- data.frame(time,Mx,My,Mz,set_nbr=x)
})
after applying some function I have output like this.
result
time Mz set_nbr
1 27810 -1.917835e-03 1
2 28980 -1.344288e-03 1
3 28350 -3.426615e-05 1
4 27900 -9.934413e-04 1
5 25560 -1.016492e-02 2
6 27360 -4.790767e-03 2
7 28080 -7.062256e-04 2
8 26550 -1.171716e-04 2
9 26820 -2.495893e-03 3
10 26550 -7.397865e-03 3
11 26550 -2.574022e-03 3
12 27990 -1.575412e-02 3
My questions starts from here.
1) How to get min,middle and max values of time column, for each set_nbr ?
2) How to use evaluated set_nbr and Mz values inside of data.list?
In short;
After deciding the min,middle and max values from time column and corresponding Mz values for each set_nbr in result, I want to return back to original data.list and extract those columns of Mx, My, Mz according those of set_nbr and Mz values. Since each set_nbr actually corresponding to 600 rows, I would like to extract those defined set_nbrs family from data.list
we use time as a factor to select set_nbr. Here factor means as extraction parameter not the real factor in R command.
In addition, as you will see four set_nbr exist for each dataset but they are indeed addressing different dataset in the data.list
I'm a big advocate of using lists of data frames when appropriate, but in this case it doesn't look like there's any reason to keep them separated as different list items. Let's combine them into a single data frame.
library(dplyr)
dat = bind_rows(data.list)
Then getting your summary stats is easy:
dat %>% group_by(set_nbr) %>%
summarize(min_time = min(time),
max_time = max(time),
middle_time = median(time))
# Source: local data frame [3 x 4]
#
# set_nbr min_time max_time middle_time
# 1 1 90 54000 27045
# 2 2 90 54000 27045
# 3 3 90 54000 27045
In your sample data, time is defined the same way each time, so of course the min, median, and max are all the same.
I'd suggest, in the new question you ask about plotting, starting with the combined data frame dat.
As to your second question:
2) How to select evaluated set_nbr values inside of data.list?
Selecting a single item from a list, use double brackets
data.list[[2]]
However, with the combined data, it's just a normal column of a normal data frame so any of these will work:
dat[dat$set_nbr == 2, ]
subset(dat, set_nbr == 2)
filter(dat, set_nbr == 2)
To your clarification in comments, if you want the Mx and My values for the time and set_nbr in the results object, using my combined dat above, simply do a join: left_join(results, dat).
This should work, but I'm a little confused because in your simulated data time is numeric, but in your new text you say "we use time as a factor". If you've converted time to a factor object, this will only work if it has the same levels in each of the data frames in your data list. If not, I would recommend keeping time as numeric.
I have a data.frame that can contains N columns (N defined at runtime), and I want to get the rows within the data frame that satisfy N-1 conditions, in other words I want to get only the rows with a specific value for the first N-1 columns.
For instance if I have a data frame with four columns (A,B,C,D) and five rows:
A B C D
1 2 3 4
9 9 9 9
1 2 9 5
4 3 2 1
1 2 3 8
I would get all the rows with A==1 & B==2 & C==3, i.e:
A B C D
1 2 3 4
1 2 3 8
But as said, the data frame can be composed of any amount of rows and columns (defined at runtime), and the values of the conditions may change.
I implemented this function (simplified):
getRows<-function(dataFrame, values) {
conditions=rep(TRUE, dim(dataFrame)[1])
for (k in 1:length(values)) {
conditions=conditions&(dataFrame[,k]==values[k])
}
return(dataFrame[conditions,])
}
Of course, this assumes the values in the values vector are sorted with respect to the columns order of the data frame, and that the length of the vector is N-1.
The function works but I've the feeling that it is not really efficient to create the vector of boolean, evaluate boolean expressions in this way and so on... especially if the data frame contains many data.
Another solution that I found is:
getRows<-function(dataFrame, values) {
tmp=dataFrame
for (k in 1:length(values)) {
tmp=tmp[tmp[,k]==values[k],]
}
return(tmp)
}
Basically this 'reduces' the data frame by filtering out all the rows that not satisfy each condition. But I think this is even worst, because it creates a new data frame object for each condition (ok always smaller, but anyway...).
So my question is: is there a method to do that more efficiently?
one possibility:
# if you are only checking for equalities
f <- function(df, values){
# values must be a list with the columns names of df as names and the conditions
# if you
y <- paste(names(values), unlist(values), sep="==", collapse=" & ")
return(df[eval(parse(text=y), envir=df),])
}
l <- as.vector(1:3, "list")
names(l) <- colnames(df)[-ncol(df)]
f(df, l)
A B C D
1 1 2 3 4
5 1 2 3 8
# you can also use other conditions
f <- function(df, values){
# values must be a list with the columns names of df as names and the conditions
# if you
y <- paste(names(values), unlist(values), collapse=" & ")
return(df[eval(parse(text=y), envir=df),])
}
l <- as.vector(paste0(c("==", "<=", "=="), 1:3), "list")
names(l) <- colnames(df)[-ncol(df)]
f(df, l)
A B C D
1 1 2 3 4
5 1 2 3 8
Sometimes matrices are quicker than data.frames to operate on, so something along the lines of:
mat <- t(as.matrix(df[-ncol(df)))
boolMat <- (mat==values) # if necessary use match to reorder values to match columns of df
ind <- colSums(boolMat)==nrow(boolMat)
df[ind,]
The idea being that values will get recycled along the columns of the matrix (which are the rows of the dataframe). colSums is meant to be quicker than an apply, so the final line should be somewhat optimised compared to apply(boolMat, 2, all).
The optimal solutions will depend on the size and proportions of the data; whether the entries are all integers; and maybe what proportion of matches you get in the data. So as #droopy mentions, you'll need to benchmark. My approach involves creating a copy of the data, so if your data is already approaching memory limits, then it might struggle - but maybe then you could generate your data in matrix rather than data.frame format to save the duplication.
I need to subset my data frame, but I do not know what condition to use.
df2<-subset(df, condition )
A part of the dataframe, `df`:
state value
a 1
b 2
c 3
a 1
b 4
c 5
I count the sum of the value column for each state using : table(df$state)
I need to create a date frame where I show just the rows where the sum of the value column is bigger then a given value x.
If x is 3, I need to have in the new data frame just the rows that have the "state" column equal to b or c.
What should I replace "condition" with? How can I use : table(df$state) in the condition?
It is not clear what are you trying to do.
table(df$state) count the occurence of each state in your data, not the sum of variable "value" for each "state".You should instead use something like this:
vv <- tapply(dat$value,dat$state,sum)
vv
a b c
2 6 8
Now you can use the result within subset, to get the sum of the value column is bigger then a given value x. For example x == 3:
subset(dat,state %in% names(vv)[vv>3])
or without using `subset ( more efficient)
dat[dat$state %in% names(vv)[vv>3],]