在R中搜索数据框中的两列

Question

I have a question about searching for values in R, it is actually a bit similar to a question which was posted yesterday (as given over here: Searching a vector/data table backwards in R ) except I think my problem is a bit more complicated (and also the opposite of what I want to do), and since I'm very new to R I'm not too sure how to solve this problem.

I have a data frame similar to one given below, and I wish to find a previous index value to my current one where the Times column is different to my current time and the Midquote column does not have an NA value.

Index               Times    |    Midquote
                -----------------------------
   1            10:30:45.58  |    5.319
   2            10:30:45.93  |    5.323
   3            10:30:45.104 |    5.325
   4            10:30:45.127 |    5.322
   5            10:30:45.188 |    5.325
   6            10:30:45.188 |    NA
   7            10:30:45.212 |    NA
   8            10:30:45.231 |    5.321
   9            10:30:45.231 |    5.321

If we start at the bottom of the data frame and take this to be the 'current' time, this is found to be at index 9 and which has a Times value of 10:30:45.231 and Midquote value of 5.321 , then if I want to find the first index where the time is different to my current time, we see this is found to be index 7, which has a time of 10:30:45.212 (since index 8 has the same time). But we also see that at index 7 the Midquote value is NA so I now have to check the data frame again. Index 6 again has a different time (ie 10:30:45.188 ) but it also has an NA value again in the Midquote column, so moving up again to index 5 we see that the Times column has a different time to my current time (ie 10:30:45.188 again) and that the Midquotes value is 5.325 .

Therefore, since at index 5 the time is 10:30:45.188 (which is different to my current time which was 10:30:45.231 ) and since the Midquote value at index 5 is not NA , I wish to obtain the output '5' since it is the index value which fulfills both criteria.

My question is, is there a good way of doing this? I am sorry if this is an easy question, I am very new to R and I don't know much about working with data frames...

EDIT: I would also like to do it preferably without adding another column to the data frame (as is given in the top answer of the link I mentioned above), if that is possible

Answer 1

Working with dates is tough especially with fractional seconds. If you could convert the times to doubles it would be easier to work with. Assuming your 'Times' are in order you could use this

library(magrittr)
which(df$Times < df[9,1] & !is.na(df$Midquote)) %>% max()

The which gives a vector of the 'Index' where 'Times' are less than that in 9 AND the 'Midquote' is not NA. The %>% sends the vector to max() which gives the highest value. This is pretty inelegant, but will get the job done.

Answer 2

If I understood it correctly, please check if this is the output you are expecting.

ind<-function(t,df){
    ind<-t
    while(t>1){
       t=t-1
        if((df$Times[t]!=df$Times[ind]) && (!is.na(df$Midquote[t]))){
            return(t)
        }
    }
}
sapply((nrow(data):1),FUN = ind,data)

#[[1]]
#[1] 5

#[[2]]
#[1] 5

#[[3]]
#[1] 5

#[[4]]
#[1] 4

#[[5]]
#[1] 4

#[[6]]
#[1] 3

#[[7]]
#[1] 2

#[[8]]
#[1] 1

#[[9]]
#NULL

The output series corresponds to the associated index for your data.frame starting from the last row.

Explanation: ind takes the value of row number as the current row , while t takes value starting from ind-1 to 1. df takes the entire data.frame as input and then while loop is used to check if time and midquote value of df$Times[t] and df$Midquote[t] satisfy the required conditions. If yes they return the index else the loop continues until it reaches the first row.

Without using sapply for a particular current row:

 ind(9,df)
 [1] 5

Answer 3

Data.table solution, 1 line.

library(data.table)

dt <- data.table(Index = 1:9,
                 Times = c( '10:30:45.58', '10:30:45.93','10:30:45.104','10:30:45.127','10:30:45.188','10:30:45.188','10:30:45.212','10:30:45.231','10:30:45.231' ),
                 Midquote = c('5.319','5.323','5.325','5.322','5.325',NA,NA,'5.321','5.321')
                )

> dt[ Times != Times[.N] & !is.na(Midquote), max(Index) ]
[1] 5

EDIT

To remove the Index column you have (at least) two options

dt2 <- data.table(Times = c( '10:30:45.58', '10:30:45.93','10:30:45.104','10:30:45.127','10:30:45.188','10:30:45.188','10:30:45.212','10:30:45.231','10:30:45.231' ),
                  Midquote = c('5.319','5.323','5.325','5.322','5.325',NA,NA,'5.321','5.321'))


# Option 1 - create an id column on the fly (unfortunately data.table recalculate .I after evaluating the "where" clause... so you need to save it)
dt2[, cbind(.SD, id=.I)][ Times != Times[.N] & !is.na(Midquote), max(id) ]

# Option 2 - simply check the last position of where your condition is met
dt2[, max(which(Times != Times[.N] & !is.na(Midquote))) ]

NB You can't do nrow because you can have, say, the 1st, 2nd, and 4th records matching your condition, and nrow would give you 3, which is wrong because the 3rd row does not match.

EDIT 2 (option 3 is not correct )

dt3 <- data.table(Times = c( '10:30:45.58', '10:30:45.93','10:30:45.104','10:30:45.127','10:30:45.188','10:30:45.188','10:30:45.212','10:30:45.231','10:30:45.231' ),
                  Midquote = c('5.319','5.323', NA,'5.322','5.325', NA, NA,'5.321','5.321'))


# Option 1 - create an id column on the fly (unfortunately data.table recalculate .I after evaluating the "where" clause... so you need to save it)
dt3[, cbind(.SD, id=.I)][ Times != Times[.N] & !is.na(Midquote), max(id) ]
[1] 5

# Option 2 - simply check the last position of where your condition is met
dt3[, max(which(Times != Times[.N] & !is.na(Midquote))) ]
[1] 5

# Option 3 - good luck with this
nrow(dt3[Times != Times[.N] & !is.na(Midquote)])
[1] 4