Is there a tidyr equivalent way of doing this? What I have works, but I'm just trying to learn the tidyr functions. Spent about an hour trying to coerce chatGPT into showing me how to do it and nothing worked. I have it broken out into multiple steps just to help me troubleshoot. I will eventually recombine into one step using pipes...
Note: The dataframe itself is pretty massive with 56000 rows and 1000 columns. I don't have a way to share it so including head() output below. I'm not sure what the minimum information needed is to help.
# Step 1: Create the key columntemp1 <- BRes3 %>% mutate(key1 = paste(Group, n, sep = "-"))# Step 2: Remove unnecessary columnstemp2 <- temp1 %>% select(key1, everything(), -mean, -`mean CI`, -sdev, -UCI, -LCI, -CIR, -n, -Group)# Step 3: Transpose the data and set column namestemp3 <- setNames(data.frame(t(temp2[-1])), temp2$key1)Structure of data frame at step2:
> head(temp2[1:5]) key1 est1 est2 est3 est4X2 X1.A-2 10.799982 10.304256 10.20124 9.550119X3 X1.A-3 10.3659866666667 10.465434 11.1673413333333 10.8014426666667X4 X1.A-4 10.679331 10.781562 10.025603 10.1510665X5 X1.A-5 11.4224648 11.6025256 10.5983192 11.618744X6 X1.A-6 10.1707603333333 11.0518533333333 10.416587 10.18374X7 X1.A-7 10.590982 10.6979774285714 10.6989508571429 10.1854577142857Desired final structure:
> head(temp3[1:5]) X1.A-2 X1.A-3 X1.A-4 X1.A-5 X1.A-6est1 10.799982 10.3659866666667 10.679331 11.4224648 10.1707603333333est2 10.304256 10.465434 10.781562 11.6025256 11.0518533333333est3 10.20124 11.1673413333333 10.025603 10.5983192 10.416587est4 9.550119 10.8014426666667 10.1510665 11.618744 10.18374est5 10.800006 9.970464 11.5241455 11.6005004 10.9990463333333est6 12.30192 11.0995906666667 10.449673 10.2591216 10.7492366666667