remove

Your question mystifies me, since it looks to me like you already know the answer.
--
Sent from my phone. Please excuse my brevity.

On February 12, 2017 3:30:49 PM PST, Val <valkremk at gmail.com> wrote:

Hi Jeff and all,
How do I get the  number of unique first names   in the two data sets?

for the first one,
result2 <- DF[ 1 == err2, ]
length(unique(result2$first))




On Sun, Feb 12, 2017 at 12:42 AM, Jeff Newmiller
<jdnewmil at dcn.davis.ca.us> wrote:

The "by" function aggregates and returns a result with generally

fewer rows

than the original data. Since you are looking to index the rows in

the

original data set, the "ave" function is better suited because it

always

returns a vector that is just as long as the input vector:

# I usually work with character data rather than factors if I plan
# to modify the data (e.g. removing rows)
DF <- read.table( text=
'first  week last
Alex    1  West
Bob     1  John
Cory    1  Jack
Cory    2  Jack
Bob     2  John
Bob     3  John
Alex    2  Joseph
Alex    3  West
Alex    4  West
', header = TRUE, as.is = TRUE )

err <- ave( DF$last
          , DF[ , "first", drop = FALSE]
          , FUN = function( lst ) {
              length( unique( lst ) )
            }
          )
result <- DF[ "1" == err, ]
result

Notice that the ave function returns a vector of the same type as was

given

to it, so even though the function returns a numeric the err
vector is character.

If you wanted to be able to examine more than one other column in
determining the keep/reject decision, you could do:

err2 <- ave( seq_along( DF$first )
           , DF[ , "first", drop = FALSE]
           , FUN = function( n ) {
              length( unique( DF[ n, "last" ] ) )
             }
           )
result2 <- DF[ 1 == err2, ]
result2

and then you would have the option to re-use the "n" index to look at

other

columns as well.

Finally, here is a dplyr solution:

library(dplyr)
result3 <- (   DF
           %>% group_by( first ) # like a prep for ave or by
           %>% mutate( err = length( unique( last ) ) ) # similar to

ave

           %>% filter( 1 == err ) # drop the rows with too many last

names

           %>% select( -err ) # drop the temporary column
           %>% as.data.frame # convert back to a plain-jane data

frame

           )
result3

which uses a small set of verbs in a pipeline of functions to go from

input

to result in one pass.

If your data set is really big (running out of memory big) then you

might

want to investigate the data.table or sqlite packages, either of

which can

be combined with dplyr to get a standardized syntax for managing

larger

amounts of data. However, most people actually aren't running out of

memory

so in most cases the extra horsepower isn't actually needed.


On Sun, 12 Feb 2017, P Tennant wrote:

Hi Val,

The by() function could be used here. With the dataframe dfr:

# split the data by first name and check for more than one last name

for

each first name
res <- by(dfr, dfr['first'], function(x) length(unique(x$last)) > 1)
# make the result more easily manipulated
res <- as.table(res)
res
# first
# Alex   Bob  Cory
# TRUE FALSE FALSE

# then use this result to subset the data
nw.dfr <- dfr[!dfr$first %in% names(res[res]) , ]
# sort if needed
nw.dfr[order(nw.dfr$first) , ]

 first week last
2   Bob    1 John
5   Bob    2 John
6   Bob    3 John
3  Cory    1 Jack
4  Cory    2 Jack


Philip

On 12/02/2017 4:02 PM, Val wrote:

Hi all,
I have a big data set and want to  remove rows conditionally.
In my data file  each person were recorded  for several weeks.

Somehow

during the recording periods, their last name was misreported.

For

each person,   the last name should be the same. Otherwise remove

from

the data. Example, in the following data set, Alex was found to

have

two last names .

Alex   West
Alex   Joseph

Alex should be removed  from the data.  if this happens then I want
remove  all rows with Alex. Here is my data set

df<- read.table(header=TRUE, text='first  week last
Alex    1  West
Bob     1  John
Cory    1  Jack
Cory    2  Jack
Bob     2  John
Bob     3  John
Alex    2  Joseph
Alex    3  West
Alex    4  West ')

Desired output

       first  week last
1     Bob     1   John
2     Bob     2   John
3     Bob     3   John
4     Cory     1   Jack
5     Cory     2   Jack

Thank you in advance

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