How to convert a list into a data frame while keeping its headers in R

Question

I have a a list called mylist, in it there are 108 rows apart from the headers. I tried converting it to a dataframe and it successfully worked using the following command

df <- data.frame(matrix(unlist(mylist), nrow=108, byrow=F))

However, the headers for each matrix in my list (mylist) have not been defined in my new dataframe df.

names(df)

is

"X1" "X2" "X3" "X4" "X5" "X6" "X7" "X8" "X9" "X10" "X11" "X12" "X13" "X14" "X15" "X16" "X17" "X18" "X19" "X20" "X21" "X22" "X23" "X24" "X25" "X26" "X27" "X28" "X29" "X30" "X31" "X32" "X33" "X34" "X35" "X36"

mylist looks something like so

head(mylist[[1]])

      RAmi         MDaf
1   11.806405   -3.588567
2   7.711101    -9.721415
3   2.315104    11.217575
4   20.372999   -2.267938
5   22.279704   -1.668082
6   13.57909    20.67355



head(mylist[[2]])

          Tomi         Rahaf
    1     325          -3
    2      71          -9
    3      2           11
    4     20.999      -22
    5      22         -16
    6     139         2065

this is only for head(mylist[[i]]) when i=1 but there exist similar things for i=1,2,3, ... , 18

What I want is to put them all in one dataframe adjacently. It worked fine but I am having a problem with the heading

I hope one may help me with understanding how to do so. Thank you

Show `mylist` or a facsimile that illustrates the problem... — Frank, Oct 25 '16 at 16:28
Use a different list that is similar enough to illustrate the problem. There is no such thing as a "header" or a "row" for a list, so what you're asking is quite unclear without an example. For advice on making a reproducible example, see: http://stackoverflow.com/a/28481250/ — Frank, Oct 25 '16 at 16:36
This could be as simple as `names(df) <- names(mylist)`, and if it's not that simple then you haven't given us enough information to support further coding. — IRTFM, Oct 25 '16 at 16:38
names(mylist) is null (no reason why). I edit it based on your request — rsc05, Oct 25 '16 at 16:42
Can you show us the second element in your list? This may just need `rbind` or `cbind`. — mikeck, Oct 25 '16 at 16:56

score 0 · Accepted Answer · answered Nov 02 '16 at 15:05

It seems that there is a problem in converting a list of dataframes into one dataframe. Therefore, it would be better to work with one dataframe from the very beginning.

For example, in my case i imported the raw data

#data1=read.csv(file.choose(), header=F)
data1=read.csv

Then afterwards I started analysing my data. I then wrote a code to generate a dataframe as in

# Here we are extracting the data which we wish to have
#iterations=nrow(data1) #Check the number of rows

listOfDataFrames <- vector(mode = "list", length = 18)   # define a dataframe with length
#mylist <- list() #create an empty list
for (k in 0:17)
{
    j=18 # from 14 i.e. switzerland this is not true anymore 
    if ((k>=14) && (k<16))
    {
        f=7+j*k
        s=3+j*k

    }  else if (k<14)
    {
        s=7+j*k-4
        f=7+j*k
    } else if (k==16)
    {
        f=j*k+5
        s=f+4
    } else if (k==17)
    {
        f=304
        s=301
    }
    b= data1[,c(f,s)] 
    #mylist[[k+1]] <- b
    listOfDataFrames[[k+1]]=b
}
#df <- data.frame(matrix(unlist(mylist), nrow=108, byrow=F))

This allowed me to put all the data into one dataframe along with the headers

df=do.call("cbind", listOfDataFrames)
head(df)

How to convert a list into a data frame while keeping its headers in R

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