merger <- cbind(as.character(Date),weather1$High,weather1$Low,weather1$Avg..High,weather1$Avg.Low,sale$Scanned.Movement[a])
After cbind the data, the new DF has column names automatically V1, V2......
I want rename the column by
colnames(merger)[,1] <- "Date"
but failed. And when I use merger$V1 ,
Error in merger$V1 : $ operator is invalid for atomic vectors
You can also name columns directly in the cbind call, e.g.
cbind(date=c(0,1), high=c(2,3))
Output:
date high
[1,] 0 2
[2,] 1 3
Try:
colnames(merger)[1] <- "Date"
Example
Here is a simple example:
a <- 1:10
b <- cbind(a, a, a)
colnames(b)
# change the first one
colnames(b)[1] <- "abc"
# change all colnames
colnames(b) <- c("aa", "bb", "cc")
you gave the following example in your question:
colnames(merger)[,1]<-"Date"
the problem is the comma: colnames() returns a vector, not a matrix, so the solution is:
colnames(merger)[1]<-"Date"
If you pass only vectors to cbind() it creates a matrix, not a dataframe. Read ?data.frame.
A way of producing a data.frame and being able to do this in one line is to coerce all matrices/data frames passed to cbind into a data.frame while setting the column names attribute using setNames:
a = matrix(rnorm(10), ncol = 2)
b = matrix(runif(10), ncol = 2)
cbind(setNames(data.frame(a), c('n1', 'n2')),
setNames(data.frame(b), c('u1', 'u2')))
which produces:
n1 n2 u1 u2
1 -0.2731750 0.5030773 0.01538194 0.3775269
2 0.5177542 0.6550924 0.04871646 0.4683186
3 -1.1419802 1.0896945 0.57212043 0.9317578
4 0.6965895 1.6973815 0.36124709 0.2882133
5 0.9062591 1.0625280 0.28034347 0.7517128
Unfortunately, there is no setColNames function analogous to setNames for data frames that returns the matrix after the column names, however, there is nothing to stop you from adapting the code of setNames to produce one:
setColNames <- function (object = nm, nm) {
colnames(object) <- nm
object
}
See this answer, the magrittr package contains functions for this.
If you offer cbind a set of arguments all of whom are vectors, you will get not a dataframe, but rather a matrix, in this case an all character matrix. They have different features. You can get a dataframe if some of your arguments remain dataframes, Try:
merger <- cbind(Date =as.character(Date),
weather1[ , c("High", "Low", "Avg..High", "Avg.Low")] ,
ScnMov =sale$Scanned.Movement[a] )
It's easy just add the name which you want to use in quotes before adding
vector
a_matrix <- cbind(b_matrix,'Name-Change'= c_vector)
Related
I'm trying to add different suffixes to my data frames so that I can distinguish them after I've merge them. I have my data frames in a list and created a vector for the suffixes but so far I have not been successful.
data2016 is the list containing my 7 data frames
new_names <- c("june2016", "july2016", "aug2016", "sep2016", "oct2016", "nov2016", "dec2016")
data2016v2 <- lapply(data2016, paste(colnames(data2016)), new_names)
Your query is not quite clear. Therefore two solutions.
The beginning is the same for either solution. Suppose you have these four dataframes:
df1x <- data.frame(v1 = rnorm(50),
v2 = runif(50))
df2x <- data.frame(v3 = rnorm(60),
v4 = runif(60))
df3x <- data.frame(v1 = rnorm(50),
v2 = runif(50))
df4x <- data.frame(v3 = rnorm(60),
v4 = runif(60))
Suppose further you assemble them in a list, something akin to your data2016using mgetand ls and describing a pattern to match them:
my_list <- mget(ls(pattern = "^df\\d+x$"))
The names of the dataframes in this list are the following:
names(my_list)
[1] "df1x" "df2x" "df3x" "df4x"
Solution 1:
Suppose you want to change the names of the dataframes thus:
new_names <- c("june2016", "july2016","aug2016", "sep2016")
Then you can simply assign new_namesto names(my_list):
names(my_list) <- new_names
And the result is:
names(my_list)
[1] "june2016" "july2016" "aug2016" "sep2016"
Solution 2:
You want to add the new_names literally as suffixes to the 'old' names, in which case you would use pasteor paste0 thus:
names(my_list) <- paste0(names(my_list), "_", new_names)
And the result is:
names(my_list)
[1] "df1x_june2016" "df2x_july2016" "df3x_aug2016" "df4x_sep2016"
You could use an index number within lapply to reference both the list and your vector of suffixes. Because there are a couple steps, I'll wrap the process in a function(). (Called an anonymous function because we aren't assigning a name to it.)
data2016v2 <- lapply(1:7, function(i) {
this_data <- data2016[[i]] # Double brackets for a list
names(this_data) <- paste0(names(this_data), new_names[i]) # Single bracket for vector
this_data # The renamed data frame to be placed into data2016v2
})
Notice in the paste0() line we are recycling the term in new_names[i], so for example if new_names[i] is "june2016" and your first data.frame has columns "A", "B", and "C" then it would give you this:
> paste0(c("A", "B", "C"), "june2016")
[1] "Ajune2016" "Bjune2016" "Cjune2016"
(You may want to add an underscore in there?)
As an aside, it sounds like you might be better served by adding the "june2016" as a column in your data (like say a variable named month with "june2016" as the value in each row) and combining your data using something like bind_rows() from the dplyr package, running it "long" instead of "wide".
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."
I have searched extensively but not found an answer to this question on Stack Overflow.
Lets say I have a data frame a.
I define:
a <- NULL
a <- as.data.frame(a)
If I wanted to add a column to this data frame as so:
a$col1 <- c(1,2,3)
I get the following error:
Error in `$<-.data.frame`(`*tmp*`, "a", value = c(1, 2, 3)) :
replacement has 3 rows, data has 0
Why is the row dimension fixed but the column is not?
How do I change the number of rows in a data frame?
If I do this (inputting the data into a list first and then converting to a df), it works fine:
a <- NULL
a$col1 <- c(1,2,3)
a <- as.data.frame(a)
The row dimension is not fixed, but data.frames are stored as list of vectors that are constrained to have the same length. You cannot add col1 to a because col1 has three values (rows) and a has zero, thereby breaking the constraint. R does not by default auto-vivify values when you attempt to extend the dimension of a data.frame by adding a column that is longer than the data.frame. The reason that the second example works is that col1 is the only vector in the data.frame so the data.frame is initialized with three rows.
If you want to automatically have the data.frame expand, you can use the following function:
cbind.all <- function (...)
{
nm <- list(...)
nm <- lapply(nm, as.matrix)
n <- max(sapply(nm, nrow))
do.call(cbind, lapply(nm, function(x) rbind(x, matrix(, n -
nrow(x), ncol(x)))))
}
This will fill missing values with NA. And you would use it like: cbind.all( df, a )
You could also do something like this where I read in data from multiple files, grab the column I want, and store it in the dataframe. I check whether the dataframe has anything in it, and if it doesn't, create a new one rather than getting the error about mismatched number of rows:
readCounts = data.frame()
for(f in names(files)){
d = read.table(files[f], header=T, as.is=T)
d2 = round(data.frame(d$NumReads))
colnames(d2) = f
if(ncol(readCounts) == 0){
readCounts = d2
rownames(readCounts) = d$Name
} else{
readCounts = cbind(readCounts, d2)
}
}
if you have an empty dataframe, called for example df, in my opinion another quite simple solution is the following:
df[1,]=NA # ad a temporary new row of NA values
df[,'new_column'] = NA # adding new column, called for example 'new_column'
df = df[0,] # delete row with NAs
I hope this may help.
I have a named list whose each element is a character vector. I want to write this list into a single dataframe where I have two columns, one with the name of the character vector and the second column with each element of the character vector. Any help would be appreciated.
NewList <- lapply(names(List), function(X) data.frame(Names=X, Characters=List[[X]]))
do.call(rbind, NewList)
Maybe
data.frame(vecname = rep(names(ll), sapply(ll, length)), chars = unlist(ll))
to have each element of each list component correspond to a row in the final dataframe.
I'm wondering if stack provides the functions you need (using the example of Henrik)
ll <- list(x1 = c("a", "b", "c"), x2 = c("d", "e"))
stack(ll)
#-------
values ind
1 a x1
2 b x1
3 c x1
4 d x2
5 e x2
A very straightforward way would be to use cbind(), like this:
cbind(names(l),l)
This will result in the following data frame, assuming that l = list(a="ax", b="bx"):
l
a "a" "ax"
b "b" "bx"
Of course, you can rename the columns and rows by adjusting the values in colnames(l) and rownames(l). In this example, the string names are automatically also applied to the rownames of the resulting data frame, so, depending on what you'd like to do with your data,
cbind(l)
might suffice, resulting in
l
a "ax"
b "bx"
Hope I could help.
I couldn't find a solution for this problem online, as simple as it seems.
Here's it is:
#Construct test dataframe
tf <- data.frame(1:3,4:6,c("A","A","A"))
#Try the apply function I'm trying to use
test <- apply(tf,2,function(x) if(is.numeric(x)) mean(x) else unique(x)[1])
#Look at the output--all columns treated as character columns...
test
#Look at the format of the original data--the first two columns are integers.
str(tf)
In general terms, I want to differentiate what function I apply over a row/column based on what type of data that row/column contains.
Here, I want a simple mean if the column is numeric and the first unique value if the column is a character column. As you can see, apply treats all columns as characters the way I've written this function.
Just write a specialised function and put it within sapply... don't use apply(dtf, 2, fun). Besides, your character ain't so characterish as you may think - run getOption("stringsAsFactors") and see for yourself.
sapply(tf, class)
X1.3 X4.6 c..A....A....A..
"integer" "integer" "factor"
sapply(tf, storage.mode)
X1.3 X4.6 c..A....A....A..
"integer" "integer" "integer"
EDIT
Or even better - use lapply:
fn <- function(x) {
if(is.numeric(x) & !is.factor(x)) {
mean(x)
} else if (is.character(x)) {
unique(x)[1]
} else if (is.factor(x)) {
as.character(x)[1]
}
}
dtf <- data.frame(a = 1:3, b = 4:6, c = rep("A", 3), stringsAsFactors = FALSE)
dtf2 <- data.frame(a = 1:3, b = 4:6, c = rep("A", 3), stringsAsFactors = TRUE)
as.data.frame(lapply(dtf, fn))
a b c
1 2 5 A
as.data.frame(lapply(dtf2, fn))
a b c
1 2 5 A
I find the numcolwise and catcolwise functions from the plyr package useful here, for a syntactically simple solution:
First let's name the columns, to avoid ugly column names when doing the aggregation:
tf <- data.frame(a = 1:3,b=4:6, d = c("A","A","A"))
Then you get your desired result with this one-liner:
> cbind(numcolwise(mean)(tf), catcolwise( function(z) unique(z)[1] )(tf))
a b d
1 2 5 A
Explanation: numcolwise(f) converts its argument ( in this case f is the mean function ) into a function that takes a data-frame and applies f only to the numeric columns of the data-frame. Similarly the catcolwise converts its function argument to a function that operates only on the categorical columns.
You want to use lapply() or sapply(), not apply(). A data.frame is a list under the hood, which apply will try to convert to a matrix before doing anything. Since at least one column in your data frame is character, every other column also gets coerced to character in forming that matrix.