Is it a way I can get the data info from global environment into a summary table?
For example, I have a lot of data set named TXXX in my global environment, like
I would like to table that looks like this
Is it possible to also get all the variable list for each data using programing?
it will looks like this:
Any way I can do that by programming? Thanks.
We can use mget to get all the objects that starts with 'T' followed by 3 digit number in to a list , then loo over the list get the number of rows, 'Obs' and number of columns 'Variable'), rbind the list elements after creating the column 'Data' as the names of the list
lst1 <- lapply(mget(ls(pattern = "^T\\d{3}$")),
function(x) data.frame(Obs = nrow(x),
Variable = ncol(x)))
out <- do.call(rbind, Map(cbind, Data = names(lst1), lst1))
row.names(out) <- NULL
If we need the column names, we could use rowr to cbind the column names when the lengths are not the same
lst1 <- lapply(mget(ls(pattern = "^T\\d{3}$")), names)
library(versions)
available.versions('rowr') # // check for available version. Not in CRAN
install.versions('rowr', '1.1.2') # // install a version
library(rowr) # // load the package
do.call(cbind.fill, c(lst1, fill = NA))
Or without installing rowr
mx <- max(lengths(lst1))
do.call(cbind, lapply(lst1, `length<-`, mx))
Or using tidyverse
library(dplyr)
library(purrr)
mget(ls(pattern = '^T\\d{3}$')) %>%
map_dfr(~ tibble(Obs = nrow(.x), Variable = ncol(.x)), .id = 'Data')
Related
I have a list with dataframes:
df1 <- data.frame(id = seq(1:10), name = LETTERS[1:10])
df2 <- data.frame(id = seq(11:20), name = LETTERS[11:20])
mylist <- list(df1, df2)
I want to remove rows from each dataframe in the list based on a condition (in this case, the value stored in column id). I create an empty vector where I will store the ids:
ids_to_remove <- c()
Then I apply my function:
sapply(mylist, function(df) {
rows_above_th <- df[(df$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)
I have the following 3 data frames, each of which has columns with names. I want to combine them and retain the column names. When I use the patch I found for combining dataframes, it drops that name on any dataframes that don't have at least 2 columns. How can I retain the names?
x<-data.frame(mean(1:10))
names(x)[names(x) == 'mean.1.10.'] <- 'var.name'
y<-data.frame(1:4)
names(y)[names(y) == 'X1.4'] <- 'var.name2'
z<-data.frame(matrix(1:10,5,2))
names(z)[names(z) == 'X1'] <- 'var.name3'
names(z)[names(z) == 'X2'] <- 'var.name4'
list_datf <- list(x, y, z)
n_r <- seq_len(max(sapply(list_datf, nrow)))
NEW <- do.call(cbind, lapply(list_datf, `[`, n_r, ))
You need to include drop = FALSE in the indexing step so that the things you're binding together retain all of their dimensions. I couldn't figure out a way to do this by passing drop = FALSE as an extra argument to [, so I resorted to using an anonymous function instead.
NEW <- do.call(cbind, lapply(list_datf, function(x) x[n_r, , drop = FALSE]))
Alternatively, you could convert your components to tibbles, which (unlike data frames) never drop "unneeded" dimensions:
NEW <- do.call(cbind, lapply(list_datf, function(x) tibble::as_tibble(x)[n_r, ]))
If you want to go full tidyverse:
library(dplyr)
list_datf %>% purrr::map(~ tibble::as_tibble(.)[n_r, ]) %>% bind_cols()
df <- data.frame("date"=
1:4,"product"=c("B","B","A","A"),"subproduct"=c("1","2","x","y"),"actuals"=1:4)
#creates df1,df2,dfx,dfy
for(i in unique(df$subproduct)) {
nam <- paste("df", i, sep = ".")
assign(nam, df[df$subproduct==i,])
}
# CREATES LIST OF DATAFRAMES
# How do I make this so i don't have to manually type list(df.,df.,df.)
list_df <- list(df.1,df.2,df.x,df.y) %>%
lapply( function(x) x[(names(x) %in% c("date", "actuals"))])
# creates df1,df2,df3,df4 only dates and actuals, removes the other column names
for (i in 1:length(list_df)) {
assign(paste0("df", i), as.data.frame(list_df[[i]]))
}
For the first for loop, it creates a df object based off unique subproduct. For the list() function, I want to be able to not have to type in df.1 ... df2... etc so if I have 100 unique subproducts in my data, I wouldn't need to type this df.1, df.2,df.x,df.y,df.z,df.zzz,df. over and over again. How would I best do this (1 question)
The last for loop creates separate dataframe objects with only date and actuals will be used to create time series for each. How can I put the values of these objects into a single dataframe or a list of dfs? (2nd question)
We can use mget to return the value of object on the subset of object names from ls. The pattern matches object names that starts with 'df'followed by a.` and any alphanumeric characters
mget(ls(pattern = '^df\\.[[:alnum:]]+$'))
If the OP wanted to create those objects in a different env
new_env <- new.env()
list2env(mget(ls(pattern = '^df\\.[[:alnum:]]+$')), envir = new_env)
If we want to create new objects from scratch, do a group_split on the 'subproduct' column, set the names accordingly, and create multiple objects (list2env - not recommended though)
library(dplyr)
library(stringr)
df %>%
group_split(subproduct) %>%
setNames(str_c('df.', c(1, 2, 'x', 'y'))) %>%
list2env(.GlobalEnv)
I have multiple .csv files (mydata_1, mydata_2,...) with the same amount of columns and column names(, different row lengths if that helps finding an answer). After reading them into my environment they have the class data.frame . I was putting them all in a list and now want to select specific columns by name from all of them, resulting in in the same variable name with just the chosen columns.
mydata_1 = matrix(c(1:21), nrow=3, ncol=7,byrow = TRUE)
mydata_2 = matrix(c(1:21), nrow=3, ncol=7,byrow = TRUE)
colnames(mydata_1) = c(paste0("X","1":"7"))
colnames(mydata_2) = c(paste0("X","1":"7"))
df1 = as.data.frame(mydata_1)
df2 = as.data.frame(mydata_2)
all_data = c(df1, df2)
class(all_data)
class(df1)
for (i in all_data){
i = select(i,"X3":"X5")
}
My for command shall output the data.frames df1 and df2 with just three columns (instead of the prior seven), but when running the code an error message regarding the select command appears.
Error in UseMethod("select_") :
no applicable method for 'select_' applied to an object of class "c('integer', 'numeric')"
How can I get an working output of my new dfs?
The first issue here is that your are trying to create a list using c(df1, df2), while you have to use list(df1, df2)
Data
library(dplyr)
library(purrr)
mydata_1 = matrix(c(1:21), nrow=3, ncol=7,byrow = TRUE)
mydata_2 = matrix(c(1:21), nrow=3, ncol=7,byrow = TRUE)
colnames(mydata_1) = c(paste0("X","1":"7"))
colnames(mydata_2) = c(paste0("X","1":"7"))
df1 = as.data.frame(mydata_1)
df2 = as.data.frame(mydata_2)
all_data = list(df1 = df1, df2 = df2)
The second problem is within your loop. look, in this approach you have to create an empty list before running the loop, and then aggregate elements in each iteration.
all_data2 <- list()
for(i in 1:length(all_data)) {
all_data2[[i]] <- all_data[[i]] %>% select(X3, X4, X5)
}
try using map from purrr which is part of the tidyverse package and lead to a cleaner code with the same result.
# Down here the `.x` is replaced by each element of the list all_data
# in each iteration, ending wiht a list of two data frames
all_data2 = map(all_data, ~.x %>%
select(X3, X4, X5))
Consider base R's subset with select argument for contiguous column selection, wrapped in an lapply call. Unlike for loop, lapply does not require the bookkeeping to reassign each element back into a list:
all_data <- list(df1 = df1, df2 = df2)
all_data_sub <- lapply(all_data, function(df) subset(df, select=X3:X5))
I have the following named list output from a analysis. The reproducible code is as follows:
list(structure(c(-213.555409754509, -212.033637890131, -212.029474755074,
-211.320398316741, -211.158815833294, -210.470525157849), .Names = c("wasn",
"chappal", "mummyji", "kmph", "flung", "movie")), structure(c(-220.119433774144,
-219.186901747536, -218.743319709963, -218.088361753899, -217.338920075687,
-217.186050877079), .Names = c("crazy", "wired", "skanndtyagi",
"andr", "unveiled", "contraption")))
I want to convert this to a data frame. I have tried unlist to data frame options using reshape2, dplyr and other solutions given for converting a list to a data frame but without much success. The output that I am looking for is something like this:
Col1 Val1 Col2 Val2
1 wasn -213.55 crazy -220.11
2 chappal -212.03 wired -219.18
3 mummyji -212.02 skanndtyagi -218.74
so on and so forth. The actual out put has multiple columns with paired values and runs into many rows. I have tried the following codes already:
do.call(rbind, lapply(df, data.frame, stringsAsFactors = TRUE))
works partially provides all the character values in a column and numeric values in the second.
data.frame(Reduce(rbind, df))
didn't work - provides the names in the first list and numbers from both the lists as tow different rows
colNames <- unique(unlist(lapply(df, names)))
M <- matrix(0, nrow = length(df), ncol = length(colNames),
dimnames = list(names(df), colNames))
matches <- lapply(df, function(x) match(names(x), colNames))
M[cbind(rep(sequence(nrow(M)), sapply(matches, length)),
unlist(matches))] <- unlist(df)
M
didn't work correctly.
Can someone help?
Since the list elements are all of the same length, you should be able to stack them and then combine them by columns.
Try:
do.call(cbind, lapply(myList, stack))
Here's another way:
as.data.frame( c(col = lapply(x, names), val = lapply(x,unname)) )
How it works. lapply returns a list; two lists combined with c make another list; and a list is easily coerced to a data.frame, since the latter is just a list of vectors having the same length.
Better than coercing to a data.frame is just modifying its class, effectively telling the list "you're a data.frame now":
L = c(col = lapply(x, names), val = lapply(x,unname))
library(data.table)
setDF(L)
The result doesn't need to be assigned anywhere with = or <- because L is modified "in place."