I have multiple data frames which are individual sequences, consisting out the same columns. I need to delete all the rows after a negative value is encountered in the column "OnsetTime". So not the row of the negative value itself, but the row after that. All sequences have 16 rows in total.
I think it must be able by a loop, but I have no experience with loops in r and I have 499 data frames of which I am currently deleting the rows of a sequence one by one, like this:
sequence_6 <- sequence_6[-c(11:16), ]
sequence_7 <- sequence_7[-c(11:16), ]
sequence_9 <- sequence_9[-c(6:16), ]
Is there a faster way of doing this? An example of a sequence can be seen here example sequence
Ragarding this example, I want to delete row 7 to row 16
Data
Since the odd web configuration at work prevents me from accessing your data, I created three dataframes based on random numbers
set.seed(123); data_1 <- data.frame( value = runif(25, min = -0.1) )
set.seed(234); data_2 <- data.frame( value = runif(20, min = -0.1) )
set.seed(345); data_3 <- data.frame( value = runif(30, min = -0.1) )
First, you could create a list containing all your dataframes:
list_df <- list(data_1, data_2, data_3)
Now you can go through this list with a for loop. Since there are several steps, I find it convenient to use the package dplyr because it allows for a more readable notation:
library(dplyr)
for( i in 1:length(list_df) ){
min_row <-
list_df[[i]] %>%
mutate( id = row_number() ) %>% # add a column with row number
filter(value < 0) %>% # get the rows with negative values
summarise( min(id) ) %>% # get the first row number
as.numeric() # transform this value to a scalar (not a dataframe)
list_df[[i]] <- list_df[[i]] %>% slice(1:min_row) # get rows 1 to min_row
}
Hope it helps!
We can get the datasets into a list assuming that the object names start with 'sequence' followed by a - and one or more digits. Then use lapply to loop over the list and subset the rows based on the condition
lst1 <- lapply(mget(ls(pattern="^sequence_\\d+$")), function(x) {
i1 <- Reduce(`|`, lapply(x, `<`, 0))
#or use rowSums
#i1 <- rowSums(x < 0) > 0
i2 <- which(i1)[1]
x[seq(i2),]
}
)
data
set.seed(42)
sequence_6 <- as.data.frame(matrix(sample(-1:10, 16 *5, replace = TRUE), nrow = 16))
sequence_7 <- as.data.frame(matrix(sample(-2:10, 16 *5, replace = TRUE), nrow = 16))
sequence_9 <- as.data.frame(matrix(sample(-2:10, 16 *5, replace = TRUE), nrow = 16))
Related
I have a data frame with 20 rows, I randomly select n rows and modify them. How can I put the modified value back to the original data frame with only the modified value being different?
df<- data.frame(rnorm(n = 20, mean = 0, sd = 1))
n = 8
a<- data.frame(df[ c(1, sample(2:(nrow(df)-1), n), nrow(df) ), ])
a$changedvalue <- a[,1]*(2.5)
Now I want to replace the values of the original dataframe df with the values of a$changedvalue such that only the sampled values are changed while everything else is same in df. I tried doing something like this but it's not working.
df %>% a[order(as.numeric(rownames(a))),]
I just want to point out that in my original dataset the data are timeseries data, so maybe they can be used for the purpose.
Instead of writing data.frame(df[ c(1, sample(2:(nrow(df)-1), n), nrow(df) ), ])
You can define the rows you want to use, lets call it rows
rows <- c(1, sample(2:(nrow(df)-1), n), nrow(df) )
Now you can do
a<- data.frame(df[rows, ])
a$changedvalue <- a[,1]*(2.5)
df[rows, ] <- a$changedvalue
I'm looking for an efficient way to select rows from a data table such that I have one representative row for each unique value in a particular column.
Let me propose a simple example:
require(data.table)
y = c('a','b','c','d','e','f','g','h')
x = sample(2:10,8,replace = TRUE)
z = rep(y,x)
dt = as.data.table( z )
my objective is to subset data table dt by sampling one row for each letter a-h in column z.
OP provided only a single column in the example. Assuming that there are multiple columns in the original dataset, we group by 'z', sample 1 row from the sequence of rows per group, get the row index (.I), extract the column with the row index ($V1) and use that to subset the rows of 'dt'.
dt[dt[ , .I[sample(.N,1)] , by = z]$V1]
You can use dplyr
library(dplyr)
dt %>%
group_by(z) %%
sample_n(1)
I think that shuffling the data.table row-wise and then applying unique(...,by) could also work. Groups are formed with by and the previous shuffling trickles down inside each group:
# shuffle the data.table row-wise
dt <- dt[sample(dim(dt)[1])]
# uniqueness by given column(s)
unique(dt, by = "z")
Below is an example on a bigger data.table with grouping by 3 columns. Comparing with #akrun ' solution seems to give the same grouping:
set.seed(2017)
dt <- data.table(c1 = sample(52*10^6),
c2 = sample(LETTERS, replace = TRUE),
c3 = sample(10^5, replace = TRUE),
c4 = sample(10^3, replace = TRUE))
# the shuffling & uniqueness
system.time( test1 <- unique(dt[sample(dim(dt)[1])], by = c("c2","c3","c4")) )
# user system elapsed
# 13.87 0.49 14.33
# #akrun' solution
system.time( test2 <- dt[dt[ , .I[sample(.N,1)] , by = c("c2","c3","c4")]$V1] )
# user system elapsed
# 11.89 0.10 12.01
# Grouping is identical (so, all groups are being sampled in both cases)
identical(x=test1[,.(c2,c3)][order(c2,c3)],
y=test2[,.(c2,c3)][order(c2,c3)])
# [1] TRUE
For sampling more than one row per group check here
Updated workflow for dplyr. I added a second column v that can be grouped by z.
require(data.table)
y = c('a','b','c','d','e','f','g','h')
x = sample(2:10,8,replace = TRUE)
z = rep(y,x)
v <- 1:length(z)
dt = data.table(z,v)
library(dplyr)
dt %>%
group_by(z) %>%
slice_sample(n = 1)
I have a matrix(100*120) and I am trying to find values <=-1 in each row for every 12 columns. I have tried several times but failed. It is easy to find values which are <= -1, but I do not know how to consider for every 12 columns and store the results for each row. Thanks for any help.
set.seed(100)
Mydata <- sample(x=-3:3,size = 100*120,replace = T)
Mydata <- matrix(data = Mydata,nrow = 100,ncol = 120)
results <- which(Mydata<=-1,arr.ind = T)
You can use the apply function to apply the which function across each column for each row at a time. If I misinterpreted what you wanted, you can adjust the MARGIN argument accordingly.
# MARGIN=1 to apply across rows
dd <- apply(Mydata,MARGIN=1,function(x) which(x <= -1))
dd[1] # which columns in row 1 have a value <= -1
You can do this using a combination of apply functions and seq()
#Example Data
set.seed(100)
Mydata <- sample(x=-3:3,size = 100*120,replace = T)
Mydata <- matrix(data = Mydata,nrow = 100,ncol = 120)
#Solution:
Myseq <- sapply(0:9,function(x) seq(1,12,1) + 12*x)
sapply(1:dim(Myseq)[2], function(x) which(Mydata[,Myseq[,x]] == -1))
This results in a list with:
each subset of the list representing one of your 10 groups of 12 columns
each value under each subset representing the position in the matrix of any value in those 12 columns with a value equal to -1.
Say I have some data of the following kind:
df<-as.data.frame(matrix(rnorm(10*10000, 1, .5), ncol=10))
I want a new dataframe that keeps the 10 original columns, but for every column retains only the highest 10 and lowest 10 values. Importantly, the rows have names corresponding to id values that need to be kept in the new data frame.
Thus, the end result data.frame is gonna be of dimensions m by 10, where m is very likely to be more than 20. But for every column, I want only 20 valid values.
The only way I can think of doing this is doing it manually per column, using dplyr and arrange, grabbing the top and bottom rows, and then creating a matrix from all the individual vectors. Clearly this is inefficient. Help?
Assuming you want to keep all the rows from the original dataset, where there is at least one value satisfying your condition (value among ten largest or ten smallest in the given column), you could do it like this:
# create a data frame
df<-as.data.frame(matrix(rnorm(10*10000, 1, .5), ncol=10))
# function to find lowes 10 and highest 10 values
lowHigh <- function(x)
{
test <- x
test[!(order(x) <= 10 | order(x) >= (length(x)- 10))] <- NA
test
}
# apply the function defined above
test2 <- apply(df, 2, lowHigh)
# use the original rownames
rownames(test2) <- rownames(df)
# keep only rows where there is value of interest
finalData <- test2[apply(apply(test2, 2, is.na), 1, sum) < 10, ]
Please note that there is definitely some smarter way of doing it...
Here is the data matrix with 10 highest and 10 lowest in each column,
x<-apply(df,2,function(k) k[order(k,decreasing=T)[c(1:10,(length(k)-9):length(k))]])
x is your 20 by 10 matrix.
Your requirement of rownames is conflicting column by column, altogether you only have 20 rownames in this matrix and it can not be same for all 10 columns. Instead, here is your order matrix,
x_roworder<-apply(df,2,function(k) order(k,decreasing=T)[c(1:10,(length(k)-9):length(k))])
This will give you corresponding rows in original data matrix within each column.
I offer a couple of answers to this.
A base R implementation ( I have used %>% to make it easier to read)
ix = lapply(df, function(x) order(x)[-(1:(length(x)-20)+10)]) %>%
unlist %>% unique %>% sort
df[ix,]
This abuses the fact that data frames are lists, finds the row id satisfying the condition for each column, then takes the unique ones in order as the row indices you want to keep. This should retain any row names attached to df
An alternative using dplyr (since you mentioned it) which if I remember correctly doesn't particular like row names
# add id as a variable
df$id = 1:nrow(df) # or row names
df %>%
gather("col",value,-id) %>%
group_by(col) %>%
filter(min_rank(value) <= 10 | min_rank(desc(value)) <= 10) %>%
ungroup %>%
select(id) %>%
left_join(df)
Edited: To fix code alignment and make a neater filter
I'm not entirely sure what you're expecting for your return / output. But this will get you the appropriate indices
# example data
set.seed(41234L)
N <- 1000
df<-data.frame(id= 1:N, matrix(rnorm(10*N, 1, .5), ncol=10))
# for each column, extract ID's for top 10 and bottom 10 values
l1 <- lapply(df[,2:11], function(x,y, n) {
xy <- data.frame(x,y)
xy <- xy[order(xy[,1]),]
return(xy[c(1:10, (n-9):n),2])
}, y= df[,1], n = N)
# check:
xx <- sort(df[,2])
all.equal(sort(df[l1[[1]], 2]), xx[c(1:10, 991:1000)])
[1] TRUE
If you want an m * 10 matrix with these unique values, where m is the number of unique indices, you could do:
l2 <- do.call("c", l1)
l2 <- unique(l2)
df2 <- df[l2,] # in this case, m == 189
This doesn't 0 / NA the columns which you're not searching on for each row. But it's unclear what your question is trying to do.
Note
This isn't as efficient as using data.table since you're going to get a copy of the data in xy <- data.frame(x,y)
Benchmark
library(microbenchmark)
microbenchmark(ira= {
test2 <- apply(df[,2:11], 2, lowHigh);
rownames(test2) <- rownames(df);
finalData <- test2[apply(apply(test2, 2, is.na), 1, sum) < 10, ]
},
alex= {
l1 <- lapply(df[,2:11], function(x,y, n) {
xy <- data.frame(x,y)
xy <- xy[order(xy[,1]),]
return(xy[c(1:10, (n-9):n),2])
}, y= df[,1], n = N);
l2 <- unique(do.call("c", l1));
df2 <- df[l2,]
}, times= 50L)
Unit: milliseconds
expr min lq mean median uq max neval cld
ira 4.360452 4.522082 5.328403 5.140874 5.560295 8.369525 50 b
alex 3.771111 3.854477 4.054388 3.936716 4.158801 5.654280 50 a
I have a table, called table_wo_nas, with multiple columns, one of which is titled ID. For each value of ID there are many rows. I want to write a function that for input x will output a data frame containing the number of rows for each ID, with column headers ID and nobs respectively as below for x <- c(2,4,8).
## id nobs
## 1 2 1041
## 2 4 474
## 3 8 192
This is what I have. It works when x is a single value (ex. 3), but not when it contains multiple values, for example 1:10 or c(2,5,7). I receive the warning "In ID[counter] <- x : number of items to replace is not a multiple of replacement length". I've just started learning R and have been struggling with this for a week and have searched manuals, this site, Google, everything. Can someone help please?
counter <- 1
ID <- vector("numeric") ## contain x
nobs <- vector("numeric") ## contain nrow
for (i in x) {
r <- subset(table_wo_nas, ID %in% x) ## create subset for rows of ID=x
ID[counter] <- x ## add x to ID
nobs[counter] <- nrow(r) ## add nrow to nobs
counter <- counter + 1 } ## loop
result <- data.frame(ID, nobs) ## create data frame
In base R,
# To make a named vector, either:
tmp <- sapply(split(table_wo_nas, table_wo_nas$ID), nrow)
# OR just:
tmp <- table(table_wo_nas$ID)
# AND
# arrange into data.frame
nobs_df <- data.frame(ID = names(tmp), nobs = tmp)
Alternately, coerce the table into a data.frame directly, and rename:
nobs_df <- data.frame(table(table_wo_nas$ID))
names(nobs_df) <- c('ID', 'nobs')
If you only want certain rows, subset:
nobs_df[c(2, 4, 8), ]
There are many, many more options; these are just a few.
With dplyr,
library(dplyr)
table_wo_nas %>% group_by(ID) %>% summarise(nobs = n())
If you only want certain IDs, add on a filter:
table_wo_nas %>% group_by(ID) %>% summarise(nobs = n()) %>% filter(ID %in% c(2, 4, 8))
Seems pretty straightforward if you just use table again:
tbl <- table( table_wo_nas[ , 'ID'] )
data.frame( IDs = names(tbl), nobs= tbl)
Could also get a quick answer although with different column names using:
as.data.frame(table( table_wo_nas[ , 'ID'] ))
Try this.
x=c(2,4,8)
count_of_id=0
#df is your data frame table_wo_nas
count_of<-function(x)
{for(i in 1 : length(x))
{count_of_id[i]<-length(which(df$id==x[i])) #find out the n of rows for each unique value of x
}
df_1<-cbind(id,count_of_id)
return(df_1)
}