r bin equal deciles - r

Question

I have a dataset containing over 6,000 observations, each record having a score ranging from 0-100. Below is a sample:
+-----+-------+
| uID | score |
+-----+-------+
| 1 | 77 |
| 2 | 61 |
| 3 | 74 |
| 4 | 47 |
| 5 | 65 |
| 6 | 51 |
| 7 | 25 |
| 8 | 64 |
| 9 | 69 |
| 10 | 52 |
+-----+-------+
I want to bin them into equal deciles based upon their rank order relative to their peers within the score column with cutoffs being at every 10th percentile, as seen below:
+-----+-------+-----------+----------+
| uID | score | position% | scoreBin |
+-----+-------+-----------+----------+
| 7 | 25 | 0.1 | 1 |
| 4 | 47 | 0.2 | 2 |
| 6 | 51 | 0.3 | 3 |
| 10 | 52 | 0.4 | 4 |
| 2 | 61 | 0.5 | 5 |
| 8 | 64 | 0.6 | 6 |
| 5 | 65 | 0.7 | 7 |
| 9 | 69 | 0.8 | 8 |
| 3 | 74 | 0.9 | 9 |
| 1 | 77 | 1 | 10 |
+-----+-------+-----------+----------+
So far I've tried cut, cut2, tapply, etc. I think I'm on the right logic path, but I have no idea on how to apply them to my situation. Any help is greatly appreciated.

Answer 1

I would use ntile() in dplyr.
library(dplyr)
score<-c(77,61,74,47,65,51,25,64,69,52)
ntile(score, 10)
##[1] 10 5 9 2 7 3 1 6 8 4
scoreBin<- ntile(score, 10)

Answer 2

In base R we can use a combination of .bincode() and quantile():
df$new <- .bincode(df$score,
breaks = quantile(df$score, seq(0, 1, by = 0.1)),
include.lowest = TRUE)
# uID score new
#1 1 77 10
#2 2 61 5
#3 3 74 9
#4 4 47 2
#5 5 65 7
#6 6 51 3
#7 7 25 1
#8 8 64 6
#9 9 69 8
#10 10 52 4

Answer 3

Here is a method that uses quantile together with cut to get the bins:
df$scoreBin <- as.integer(cut(df$score,
breaks=quantile(df$score, seq(0,1, .1), include.lowest=T)))
as.integer coerces the output of cut (which is a factor) into the underlying integer.
One way to get the position percent is to use rank:
df$position <- rank(df$score) / nrow(df)