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I have a dataframe and want the number variable to be four digits long, in order to do this I need to add between 1-3 leading zeroes, the method I chose to do this is the sprintf function, as it is immaterial that the number is converted to character class. Unfortunately the results are not coming out in the order I want

The test data frame is made as follows and the leading 0 column added on as a third column to allow easy comparison. As can be seen by running the code the order that the leading zero numbers are pasted in does not correspond to the original number order

test <- as.data.frame(cbind(letters,seq(from=1, to=26)))
test[,3]<-sprintf("%04d", test[,2])

by rearranging the data frame order alphabetically by classing the original number column as characters, the sprintf number are now in ascending order although the number series is not.

test.two <- as.data.frame(cbind(letters,seq(from=1, to=26)))
test.two <- test.two[i <-order(as.character(test.two[,2])),]
test.two[,3]<-sprintf("%04d", test.two[,2])

I can create the desired data set by Frankensteining it togther.

test.three <- as.data.frame(cbind(letters,seq(from=1, to=26)))
test.three[,3]<-test.two[,3]

However I would like to know what I am doing wrong and what method would give me the result I expected to get from what I thought was a simple operation!

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1 回答 1

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这是因为第二列是一个因素。

test <- as.data.frame(cbind(letters,seq(from=1, to=26)))
sapply(test, class)
##  letters       V2 
## "factor" "factor" 
test[,3]<-sprintf("%04d", test[,2])

as.numeric(test$V2)
##  [1]  1 12 20 21 22 23 24 25 26  2  3  4  5  6  7  8  9 10 11 13 14 15 16 17 18
## [26] 19

test$V2 <- as.integer(as.character(test$V2))
test[,4]<-sprintf("%04d", test[,2])

##   letters V2   V3   V4
## 1       a  1 0001 0001
## 2       b  2 0012 0002
## 3       c  3 0020 0003
## 4       d  4 0021 0004
## 5       e  5 0022 0005
## 6       f  6 0023 0006
于 2014-02-27T11:38:20.710 回答