Help with data management

You can also combine the data frames into a single one and use xtabs:

ID <- names(mylist)
mylist <- Map(data.frame, mylist, dfn=ID)
mydf <- do.call(rbind, mylist)
mydf$Family <- factor(mydf$Family, levels=sort(levels(mydf$Family)))
xtabs(Hits~Family+dfn, mydf)
#       dfn
# Family  A  B  C
#      a  0  3  0
#      c  1  1  0
#      d  2  0  0
#      e  3  0  0
#      f  0  4  5
#      o  0  0  4
#      q  0  0 10


-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352




-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Jim Lemon
Sent: Thursday, February 23, 2017 6:00 PM
To: Andr? Luis Neves <andrluis at ualberta.ca>; r-help mailing list <
r-help at r-project.org>
Subject: Re: [R] Help with data management

Hi Andre,
As far as I am aware, merges can only be accomplished between two data
frames, so I think you would have to do it one by one. It is probably
possible to program this to operate on your list of data frames, but I
suspect that it would take as much time as a bit of copying and
pasting. If your data is being extracted from an external database, it
may be possible to perform the operation in SQL, I don't have the time
to work that out at the moment.

Jim


On Fri, Feb 24, 2017 at 10:53 AM, Andr? Luis Neves <andrluis at ualberta.ca>
wrote:

Hi, Jim:

Your code worked great, but I have 48 dataframes. After merging A and B

in

D, you merged C in D. In this case, do I need to add them one by one

until

getting the 48 dataframes merged in one?

Thank you for your great help.

Andre

On Thu, Feb 23, 2017 at 4:24 PM, Jim Lemon <drjimlemon at gmail.com> wrote:

Hi Andre,
This might do it:

A<-data.frame(c("c", "d", "e"),4.4:6.8,c(1,2,3))
colnames(A) <- c ("Family", "NormalizedCount", "Hits")
B<-data.frame(c("c", "f", "a"),c(3.2,6.4, 4.4), c(1,4,3))
colnames(B) <- c ("Family", "NormalizedCount", "Hits")
C<-data.frame(c("q", "o", "f"),c(7.2,9.4, 41.4), c(10,4,5))
colnames(C) <- c ("Family", "NormalizedCount", "Hits")
keepcols<-c("Family","Hits")
D<-merge(A[,keepcols],B[,keepcols],by="Family",all=TRUE)
D<-merge(D,C[,keepcols],by="Family",all=TRUE)
D[,2:4]<-sapply(D[,-1],function(x) { x[is.na(x)]<-0; x })
names(D)<-c("Family","A","B","C")

Jim


On Fri, Feb 24, 2017 at 9:37 AM, Andr? Luis Neves <andrluis at ualberta.ca

wrote:

Dear R users,

I have the following dataframes (A, B, and C) stored in a list:

A= data.frame(c("c", "d", "e"),4.4:6.8,c(1,2,3))
colnames(A) <- c ("Family", "NormalizedCount", "Hits")
A


B= data.frame(c("c", "f", "a"),c(3.2,6.4, 4.4), c(1,4,3))
colnames(B) <- c ("Family", "NormalizedCount", "Hits")
B


C= data.frame(c("q", "o", "f"),c(7.2,9.4, 41.4), c(10,4,5))
colnames(C) <- c ("Family", "NormalizedCount", "Hits")
C

mylist <- list(A=A,B=B,C=C)
mylist


My idea is to merge the three dataframes into another dataframe (let's
name
it: 'D')  with a structure in which the rows are the Families and
columns
the "Hits" of each family detected in the dataframes A, B, and C. If a
given 'Family' does NOT have a 'Hit' in the dataframe we need to

assign

number 0 to it.

The dataframe 'D' would need to be populated as follows:


Family                                                      A
       B                                      C
c 1 1 0
d 2 0 0
e 3 0 0
f 0 4 5
a 0 3 0
q 0 0 10
o 0 0 4


Thank you very much for your great help,



--
Andre

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--
Andre