I have a data frame like the one you see here.
DRSi TP DOC DN date Turbidity Anions
158 5.9 3371 264 14/8/06 5.83 2246.02
217 4.7 2060 428 16/8/06 6.04 1632.29
181 10.6 1828 219 16/8/06 6.11 1005.00
397 5.3 1027 439 16/8/06 5.74 314.19
2204 81.2 11770 1827 15/8/06 9.64 2635.39
307 2.9 1954 589 15/8/06 6.12 2762.02
136 7.1 2712 157 14/8/06 5.83 2049.86
1502 15.3 4123 959 15/8/06 6.48 2648.12
1113 1.5 819 195 17/8/06 5.83 804.42
329 4.1 2264 434 16/8/06 6.19 2214.89
193 3.5 5691 251 17/8/06 5.64 1299.25
1152 3.5 2865 1075 15/8/06 5.66 2573.78
357 4.1 5664 509 16/8/06 6.06 1982.08
513 7.1 2485 586 15/8/06 6.24 2608.35
1645 6.5 4878 208 17/8/06 5.96 969.32
Before I got here i used the following code to remove those columns that had no values at all or some NA's.
rem = NULL
for(col.nr in 1:dim(E.3)[2]){
if(sum(is.na(E.3[, col.nr]) > 0 | all(is.na(E.3[,col.nr])))){
rem = c(rem, col.nr)
}
}
E.4 <- E.3[, -rem]
Now I need to remove the "date" column but not based on its column name, rather based on the fact that it's a character string.
I've seen here ( Remove an entire column from a data.frame in R ) already how to simply set it to NULL and other options but I want to use a different argument.
First use is.character
to find all columns with class character
. However, make sure that your date is really a character
, not a Date
or a factor
. Otherwise use is.Date
or is.factor
instead of is.character
.
Then just subset the columns that are not characters in the data.frame, eg
df[, !sapply(df, is.character)]
I was having a similar problem but the answer above isn't resolve it for a Date columns (that's what I needed), so I've found another solution:
df[,-grep ("Date|factor|character", sapply (df, class))]
Will return you your df without Date, character and factor columns.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.