xts - Delete specific rows without transform to other format
Question: How can I delete specific rows in xts, without the need of transform to other formats. The selected row to be deleted, will be based on the row number, or [column.one] but not the timestamp.
This is what I have tested so far:
Send row 4, column 1, to NULL
myxts[4,1] <- NULL
Result: Error in NextMethod(.Generic) :
number of items to replace is not a multiple of replacement length
Add value to NA, then delete all NA with [na.omit]
xts3[1,1] <- NA # Adds NA
na.omit(xts3)
Result: Does remove the NA rows but does not solve the straightforward way of deleting a row.
This is my test xts:
##########
# Test xts
##########
dates <- as.POSIXct( # Construct the dates to be used.
c(
"2013-07-24 09:01:00",
"2013-07-24 09:02:00",
"2013-07-24 09:03:00"
)
)
column.one <- c(1,2,3) # Data in column.one.
data <- data.frame(column.one) # Create a dataframe.
xts3 <- xts(x=data, order.by=dates) # Create xts based on dataframe.
r xts
asked Nov 22 at 12:33
Toolbox
560310
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Question: How can I delete specific rows in xts, without the need of transform to other formats. The selected row to be deleted, will be based on the row number, or [column.one] but not the timestamp.
This is what I have tested so far:
Send row 4, column 1, to NULL
myxts[4,1] <- NULL
Result: Error in NextMethod(.Generic) :
number of items to replace is not a multiple of replacement length
Add value to NA, then delete all NA with [na.omit]
xts3[1,1] <- NA # Adds NA
na.omit(xts3)
Result: Does remove the NA rows but does not solve the straightforward way of deleting a row.
This is my test xts:
##########
# Test xts
##########
dates <- as.POSIXct( # Construct the dates to be used.
c(
"2013-07-24 09:01:00",
"2013-07-24 09:02:00",
"2013-07-24 09:03:00"
)
)
column.one <- c(1,2,3) # Data in column.one.
data <- data.frame(column.one) # Create a dataframe.
xts3 <- xts(x=data, order.by=dates) # Create xts based on dataframe.
r xts
asked Nov 22 at 12:33
Toolbox
560310
add a comment |
Question: How can I delete specific rows in xts, without the need of transform to other formats. The selected row to be deleted, will be based on the row number, or [column.one] but not the timestamp.
This is what I have tested so far:
Send row 4, column 1, to NULL
myxts[4,1] <- NULL
Result: Error in NextMethod(.Generic) :
number of items to replace is not a multiple of replacement length
Add value to NA, then delete all NA with [na.omit]
xts3[1,1] <- NA # Adds NA
na.omit(xts3)
Result: Does remove the NA rows but does not solve the straightforward way of deleting a row.
This is my test xts:
##########
# Test xts
##########
dates <- as.POSIXct( # Construct the dates to be used.
c(
"2013-07-24 09:01:00",
"2013-07-24 09:02:00",
"2013-07-24 09:03:00"
)
)
column.one <- c(1,2,3) # Data in column.one.
data <- data.frame(column.one) # Create a dataframe.
xts3 <- xts(x=data, order.by=dates) # Create xts based on dataframe.
r xts
asked Nov 22 at 12:33
Toolbox
560310
Question: How can I delete specific rows in xts, without the need of transform to other formats. The selected row to be deleted, will be based on the row number, or [column.one] but not the timestamp.
This is what I have tested so far:
Send row 4, column 1, to NULL
myxts[4,1] <- NULL
Result: Error in NextMethod(.Generic) :
number of items to replace is not a multiple of replacement length
Add value to NA, then delete all NA with [na.omit]
xts3[1,1] <- NA # Adds NA
na.omit(xts3)
Result: Does remove the NA rows but does not solve the straightforward way of deleting a row.
This is my test xts:
##########
# Test xts
##########
dates <- as.POSIXct( # Construct the dates to be used.
c(
"2013-07-24 09:01:00",
"2013-07-24 09:02:00",
"2013-07-24 09:03:00"
)
)
column.one <- c(1,2,3) # Data in column.one.
data <- data.frame(column.one) # Create a dataframe.
xts3 <- xts(x=data, order.by=dates) # Create xts based on dataframe.
r xts
r xts
asked Nov 22 at 12:33
Toolbox
560310
asked Nov 22 at 12:33
Toolbox
560310
asked Nov 22 at 12:33
Toolbox
560310
asked Nov 22 at 12:33
Toolbox
560310
asked Nov 22 at 12:33
Toolbox
560310
560310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The following lines result in equal output.
xts3[-2, ]
xts3[index(xts3) != index(xts3[xts3$column.one == 2])]
column.one
2013-07-24 09:01:00 1
2013-07-24 09:03:00 3
But for xts / zoo timeseries it is better and safer to work with the indexes as this leads to a finer control of what you want / can achieve with them.
answered Nov 22 at 13:13
phiver
12.4k92634
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The following lines result in equal output.
xts3[-2, ]
xts3[index(xts3) != index(xts3[xts3$column.one == 2])]
column.one
2013-07-24 09:01:00 1
2013-07-24 09:03:00 3
But for xts / zoo timeseries it is better and safer to work with the indexes as this leads to a finer control of what you want / can achieve with them.
answered Nov 22 at 13:13
phiver
12.4k92634
add a comment |
The following lines result in equal output.
xts3[-2, ]
xts3[index(xts3) != index(xts3[xts3$column.one == 2])]
column.one
2013-07-24 09:01:00 1
2013-07-24 09:03:00 3
But for xts / zoo timeseries it is better and safer to work with the indexes as this leads to a finer control of what you want / can achieve with them.
answered Nov 22 at 13:13
phiver
12.4k92634
add a comment |
The following lines result in equal output.
xts3[-2, ]
xts3[index(xts3) != index(xts3[xts3$column.one == 2])]
column.one
2013-07-24 09:01:00 1
2013-07-24 09:03:00 3
But for xts / zoo timeseries it is better and safer to work with the indexes as this leads to a finer control of what you want / can achieve with them.
answered Nov 22 at 13:13
phiver
12.4k92634
The following lines result in equal output.
xts3[-2, ]
xts3[index(xts3) != index(xts3[xts3$column.one == 2])]
column.one
2013-07-24 09:01:00 1
2013-07-24 09:03:00 3
But for xts / zoo timeseries it is better and safer to work with the indexes as this leads to a finer control of what you want / can achieve with them.
answered Nov 22 at 13:13
phiver
12.4k92634
answered Nov 22 at 13:13
phiver
12.4k92634
answered Nov 22 at 13:13
phiver
12.4k92634
answered Nov 22 at 13:13
phiver
12.4k92634
12.4k92634
add a comment |
add a comment |
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