I have a list with dataframes:
df1 <- data.frame(id = seq(1:10), name = LETTERS[1:10])
df2 <- data.frame(id = seq(11:20), name = LETTERS[11:20])
mylist <- list(df1, df2)
I want to remove rows from each dataframe in the list based on a condition (in this case, the value stored in column id). I create an empty vector where I will store the ids:
ids_to_remove <- c()
Then I apply my function:
sapply(mylist, function(df) {
rows_above_th <- df[(df$id > 8),] # select the rows from each df above a threshold
a <- rows_above_th$id # obtain the ids of the rows above the threshold
ids_to_remove <- append(ids_to_remove, a) # append each id to the vector
},
simplify = T
)
However, with or without simplify = T, this returns a matrix, while my desired output (ids_to_remove) would be a vector containing the ids, like this:
ids_to_remove <- c(9,10,9,10)
Because lastly I would use it in this way on single dataframes:
for(i in 1:length(ids_to_remove)){
mylist[[1]] <- mylist[[1]] %>%
filter(!id == ids_to_remove[i])
}
And like this on the whole list (which is not working and I don´t get why):
i = 1
lapply(mylist,
function(df) {
for(i in 1:length(ids_to_remove)){
df <- df %>%
filter(!id == ids_to_remove[i])
i = i + 1
}
} )
I get the errors may be in the append part of the sapply and maybe in the indexing of the lapply. I played around a bit but couldn´t still find the errors (or a better way to do this).
EDIT: original data has 70 dataframes (in a list) for a total of 2 million rows
If you are using sapply/lapply you want to avoid trying to change the values of global variables. Instead, you should return the values you want. For example generate a vector if IDs to remove for each item in the list as a list
ids_to_remove <- lapply(mylist, function(df) {
rows_above_th <- df[(df$id > 8),] # select the rows from each df above a threshold
rows_above_th$id # obtain the ids of the rows above the threshold
})
And then you can use that list with your data list and mapply to iterate the two lists together
mapply(function(data, ids) {
data %>% dplyr::filter(!id %in% ids)
}, mylist, ids_to_remove, SIMPLIFY=FALSE)
Using base R
Map(\(x, y) subset(x, !id %in% y), mylist, ids_to_remove)
So, I have a list of strings named control_for. I have a data frame sampleTable with some of the columns named as strings from control_for list. And I have a third object dge_obj (DGElist object) where I want to append those columns. What I wanted to do - use lapply to loop through control_for list, and for each string, find a column in sampleTable with the same name, and then add that column (as a factor) to a DGElist object. For example, for doing it manually with just one string, it looks like this, and it works:
group <- as.factor(sampleTable[,3])
dge_obj$samples$group <- group
And I tried something like this:
lapply(control_for, function(x) {
x <- as.factor(sampleTable[, x])
dge_obj$samples$x <- x
}
Which doesn't work. I guess the problem is that R can't recognize addressing columns like this. Can someone help?
Here are two base R ways of doing it. The data set is the example of help("DGEList") and a mock up data.frame sampleTable.
Define a vector common_vars of the table's names in control_for. Then create the new columns.
library(edgeR)
sampleTable <- data.frame(a = 1:4, b = 5:8, no = letters[21:24])
control_for <- c("a", "b")
common_vars <- intersect(control_for, names(sampleTable))
1. for loop
for(x in common_vars){
y <- sampleTable[[x]]
dge_obj$samples[[x]] <- factor(y)
}
2. *apply loop.
tmp <- sapply(sampleTable[common_vars], factor)
dge_obj$samples <- cbind(dge_obj$samples, tmp)
This code can be rewritten as a one-liner.
Data
set.seed(2021)
y <- matrix(rnbinom(10000,mu=5,size=2),ncol=4)
dge_obj <- DGEList(counts=y, group=rep(1:2,each=2))
I need to add several thousand columns to a data frame. Currently, I have a list of 93 lists, where each of the embedded lists contains 4 data frames, each with 19 variables. I want to add each column of all those data frames to an outside file. My code looks like:
vars <- c('tmin_F','tavg_F','tmax_F','pp','etr_grass','etr_alfalfa','vpd','rhmin','rhmax','dtr_F','us','shum','pp_def_grass','pp_def_alfalfa','rw_tot','fdd28_F0','fdd32_F0','fdd35_F0',
'fdd356_F0','fdd36_F0','fdd38_F0','fdd39_F0','fdd392_F0','fdd40_F0','fdd41_F0','fdd44_F0','fdd45_F0','fdd464_F0','fdd48_F0','fdd50_F0','fdd52_F0','fdd536_F0','fdd55_F0',
'fdd57_F0','fdd59_F0','fdd60_F0','fdd65_F0','fdd70_F0','fdd72_F0','hdd40_F0','hdd45_F0','hdd50_F0','hdd55_F0','hdd57_F0','hdd60_F0','hdd65_F0','hdd45_F0',
'cdd45_F0','cdd50_F0','cdd55_F0','cdd57_F0','cdd60_F0','cdd65_F0','cdd70_F0','cdd72_F0',
'gdd32_F0','gdd35_F0','gdd356_F0','gdd38_F0','gdd39_F0','gdd392_F0','gdd40_F0','gdd41_F0','gdd44_F0','gdd45_F0',
'gdd464_F0','gdd48_F0','gdd50_F0','gdd52_F0','gdd536_F0','gdd55_F0','gdd57_F0','gdd59_F0','gdd60_F0','gdd65_F0','gdd70_F0','gdd72_F0',
'gddmod_32_59_F0','gddmod_32_788_F0','gddmod_356_788_F0','gddmod_392_86_F0','gddmod_41_86_F0','gddmod_464_86_F0','gddmod_48_86_F0','gddmod_50_86_F0','gddmod_536_95_F0',
'sdd77_F0','sdd86_F0','sdd95_F0','sdd97_F0','sdd99_F0','sdd104_F0','sdd113_F0')
windows <- c(15,15,15,29,29,29,15,15,15,15,29,29,29,29,15,rep(15,78))
perc_list <- c('obs','smoothed_obs','windowed_obs','smoothed_windowed_obs')
percs <- c('00','02','05','10','20','25','30','33','40','50','60','66','70','75','80','90','95','98','100')
vcols <- seq(1,19,1)
for (v in 1:93){
for (pl in 1:4){
for (p in 1:19){
normals_1981_2010 <- normals_1981_2010 %>% mutate(!!paste0(vars[v],'_daily',perc_list[pl],'_perc',percs[p]) := percents[[v]][[pl]][,vcols[p]])}}
print(v)}
The code starts fast, but very quickly slows to a crawl as the outside data frame grows in size. I didn't realize this would be problem. How do I add all these extra columns efficiently? Is there a better way to do this than by using mutate? I've tried add_column, but that does not work. Maybe it doesn't like the loop or something.
Your example is not reproducible as is (the object normals_1981_2010 doesn't exist but is called within the loop, so I am unsure I understood your question.
If I did though, this should work:
First, I am reproducing your dataset structure, except that instead of 93 list, I set it up to have 5, instead of 4 nested tables within, I set it up to have 3 tables, and instead of each tables having 19 columns, I set them up to have 3 columns.
df_list <- vector("list", 5) # Create an empty list vector, then fill it in.
for(i in 1:5) {
df_list[[i]] <- vector("list", 3)
for(j in 1:3) {
df_list[[i]][[j]] <- data.frame(a = 1:12,
b = letters[1:12],
c = month.abb[1:12])
colnames(df_list[[i]][[j]]) <- paste0(colnames(df_list[[i]][[j]]), "_nest_", i, "subnest_", j)
}
}
df_list # preview the structure.
Then, answering your question:
# Now, how to bind everything together:
df_out <- vector("list", 5)
for(i in 1:5) {
df_out[[i]] <- bind_cols(df_list[[i]])
}
# Final step
df_out <- bind_cols(df_out)
ncol(df_out) # Here I have 5*3*3 = 45 columns, but you will have 93*4*19 = 7068 columns
# [1] 45
I'm fairly new to R and I was wondering if someone could help me?
I have a list of identical data frames (df1, df2, ..., df9) and I'm trying to rename one of the columns, 'value', in all the data frames to be 'value_dataframename'- the renamed column in all 9 data frames should be value_df1 in df1, value_df2 in df2, ..., value_df9 in df9.
Any help would be much appreciated!
Below code with example list (auto.list) that does what you want. Run it to check.
To use it for your list:
skip the code till the your.list <- ... line,
save your list as your.list object,
assign to term your "value".
auto.list <- list()
for (i in seq_len(10)) {
auto.list[[i]] <- data.frame("a" = 1:i, "value" = sample(letters, i))
names(auto.list)[i] <- paste0("df", i)
}
your.list <- auto.list # assign to your.list your own list
term <- "value" # assign your own "value"
for (i in seq_along(your.list)) {
colnames(your.list[[i]])[colnames(your.list[[i]]) == term] <- paste0(term, "_", names(your.list)[i])
}
Try this out:
## these two are my sample data frames for this example
df_1 <- data.frame(first = rbinom(10,size = 2,prob = 0.3), second = rnorm(10))
df_2 <- data.frame(first = rbinom(10,size = 2,prob = 0.3), second = rnorm(10))
# R stores data frames as list, so you can retrieve all your data frames thus:
all_df_names = ls.str(mode = "list")
# to check: all_df_names[1] - the first element - will give you "df_1", which is the name of the first data frame
# be careful though - 'ls.str(mode = "list")' will pick ALL the lists currently in your environment
# if you don't want to use this ls method, it might be wiser to manually create a variable 'all_df_names' and put all your data frame names there yourself.
# rename
for(i in 1:length(all_df_names)) {
# get the actual content via its variable name, and store it in a temporary variable 'x'
x = get(all_df_names[i])
# rename the column you want
names(x)[2] = paste0(names(x)[2], "_", i) # this will replace the column with the previous name plus a '_' and the current iteration
# resave that dataframe, with the new content
assign(all_df_names[i], x)
}
# to remove variables we no longer need when done:
# rm(x, i)
# confirm
# names(df_1) = "first" "second_1"
# names(df_2) = "first" "second_2"
I have a for loop that generates a dataframe every time it loops through. I am trying to create a list of data frames but I cannot seem to figure out a good way to do this.
For example, with vectors I usually do something like this:
my_numbers <- c()
for (i in 1:4){
my_numbers <- c(my_numbers,i)
}
This will result in a vector c(1,2,3,4). I want to do something similar with dataframes, but accessing the list of data frames is quite difficult when i use:
my_dataframes <- list(my_dataframes,DATAFRAME).
Help please. The main goal is just to create a list of dataframes that I can later on access dataframe by dataframe. Thank you.
I'm sure you've noticed that list does not do what you want it to do, nor should it. c also doesn't work in this case because it flattens data frames, even when recursive=FALSE.
You can use append. As in,
data_frame_list = list()
for( i in 1:5 ){
d = create_data_frame(i)
data_frame_list = append(data_frame_list,)
}
Better still, you can assign directly to indexed elements, even if those elements don't exist yet:
data_frame_list = list()
for( i in 1:5 ){
data_frame_list[[i]] = create_data_frame(i)
}
This applies to vectors, too. But if you want to create a vector c(1,2,3,4) just use 1:4, or its underlying function seq.
Of course, lapply or the *lply functions from plyr are often better than looping depending on your application.
Continuing with your for loop method, here's a little example of creating and accessing.
> my_numbers <- vector('list', 4)
> for (i in 1:4) my_numbers[[i]] <- data.frame(x = seq(i))
And we can access the first column of each data frame with,
> sapply(my_numbers, "[", 1)
# $x
# [1] 1
#
# $x
# [1] 1 2
#
# $x
# [1] 1 2 3
#
# $x
# [1] 1 2 3 4
Other ways of accessing the data is my_numbers[[1]] for the first data set,
lapply(my_numbers, "[", 1,) to access the first row of each data frame, etc.
You can use operator [[ ]] for this purpose.
l <- list()
df1 <- data.frame(name = 'df1', a = 1:5 , b = letters[1:5])
df2 <- data.frame(name = 'df2', a = 6:10 , b = letters[6:10])
df3 <- data.frame(name = 'df3', a = 11:20 , b = letters[11:20])
df <- rbind(df1,df2,df3)
for(df_name in unique(df$name)){
l[[df_name]] <- df[df$name == df_name,]
}
In this example, there are three separate data frames and in order to store them
in a list using a for loop, we place them in one. Using the operator [[ we can even name the data frame in the list as we want and store it in the list normally.