How to convert row names to column names and bind it by order in r

Question

I am working on a data set: Annual Returns by Ticker

and I want to convert to : Result Matrix

I used the code:

returns.df <- returns %>% spread(key = DATE, value = RETURN)

and its is showing wrong order: enter image description here

Just want to check: your complaint is the order of the columns in the result? — Gregor Thomas, Apr 05 '19 at 20:45
Images do not provide a reproducible way to run the code. Please provide a reproducible example, see here https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example?rq=1 — cropgen, Apr 05 '19 at 20:45

score 0 · Accepted Answer · answered Apr 05 '19 at 20:49

Assuming your long-form data starts with the rows in the order you want, try this:

month_order = unique(returns)
returns.df <- returns %>%
  spread(key = DATE, value = RETURN) %>% 
  select(c("TICKER", month_order))

If your data doesn't begin in the right order, append a year and convert it to a Date class object. Then you can sort it to the right order and use the method above.

score 0 · Answer 2 · edited Apr 05 '19 at 21:01

0

I figured it out my writing these multiple lines:

column <- unique(returns$DATE)
ret <- matrix(returns$RETURN,nrow = 22,ncol = 60)
row <- unique(returns$TICKER)
rownames(ret) <- row
colnames(ret) <- column
ret

How about this one?

edited Apr 05 '19 at 21:01

Jilber Urbina

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answered Apr 05 '19 at 20:54

Ritwick Dutta

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How to convert row names to column names and bind it by order in r

2 Answers2