Hi John, Thanks. As you said, I want the "Site" values to become individual variable names whose value is in the "x" column: Output should look exactly like: Year G101 G102 2001 33.1 34 Thanks, Atem.
On Sunday, August 3, 2014 4:48 PM, John McKown <john.archie.mckown at gmail.com> wrote:
On Sun, Aug 3, 2014 at 5:43 PM, John McKown
<john.archie.mckown at gmail.com> wrote:
On Sun, Aug 3, 2014 at 5:12 PM, Zilefac Elvis <zilefacelvis at yahoo.com> wrote:
Dear All, I have a List in R and would like to convert it to data.frame. Below is a reproducible example. I can do something like: x1<-do.call(cbind.data.frame, lst3) #OR x1<-as.data.frame(lst3). However, my output looks like this: Year Site? x? Year Site? x 2001 G101 33.1 2001 G102 34 I would like to have as my output: Year? G101? G102 2001? 33.1? 34 Thanks Atem.
<snip> Try rbind_all in the dplyr package. output <- rbind_all(as.list(lst3)); Example transcript:
lst3 <- list(structure(list(Year = 2001L, Site = structure(1L, .Label = "G101", class = "factor"),
+? ? x = 33.1), .Names = c("Year", "Site", "x"), row.names = c(NA,
+ -1L), class = "data.frame"), structure(list(Year = 2001L, Site =
structure(1L, .Label = "G102", class = "factor"),
+? ? x = 34), .Names = c("Year", "Site", "x"), row.names = c(NA,
+ -1L), class = "data.frame"))
library(dplyr); rbind_all(lst3);
? Year Site? ? x 1 2001 G101 33.1 2 2001 G102 34.0 Warning message: In rbind_all(lst3) : Unequal factor levels: coercing to character
OOPS. Didn't got far enough. You want the "Site" values to become individual variable names whose value is in the "x" column. Some days I need more naps. But I think that this is the starting point. More later. Sorry.
-- There is nothing more pleasant than traveling and meeting new people! Genghis Khan Maranatha! <>< John McKown
There is nothing more pleasant than traveling and meeting new people! Genghis Khan Maranatha! <>< John McKown