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Referencing column names in data table =:
Select / assign to data.table when variable names are stored in a character vectorHow to sort a dataframe by multiple column(s)Referencing a column vector in a data frame within a loopDrop data frame columns by nameChanging column names of a data frameLoop through columns in a data.table and transform those columnsdata.table vs dplyr: can one do something well the other can't or does poorly?dplyr - mutate: use dynamic variable namesget numeric column values in data table in rIn R produce a running count in column based on previous column with numeric valuesSubtract two data.tables
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
I'm running a loop in which two numeric columns are created based on some data. These two column names are generated by the part of the loop they are in. So for example, the first two columns are df$loop1.1 and df$loop1.2 and then in the following loop they'd be df$loop2.1 and df$loop2.2
I need to create a third column that is based off of those two columns like so:
df[, divisioncol := loop1.1 / loop1.2]
However, I'm getting the error
non-numeric argument to binary operator
Any suggestions?
Solved:
df[, divisioncol := get(loop1.1)/get(loop2.2)]
r data.table
asked Mar 25 at 20:46
JustinJustin
223 bronze badges
add a comment |
I'm running a loop in which two numeric columns are created based on some data. These two column names are generated by the part of the loop they are in. So for example, the first two columns are df$loop1.1 and df$loop1.2 and then in the following loop they'd be df$loop2.1 and df$loop2.2
I need to create a third column that is based off of those two columns like so:
df[, divisioncol := loop1.1 / loop1.2]
However, I'm getting the error
non-numeric argument to binary operator
Any suggestions?
Solved:
df[, divisioncol := get(loop1.1)/get(loop2.2)]
r data.table
asked Mar 25 at 20:46
JustinJustin
223 bronze badges
1
If you are passing the column names, maybe wrap inget, likeget(loop1.1) / get(loop1.2)stackoverflow.com/q/12391950
– Frank
Mar 25 at 20:58
add a comment |
I'm running a loop in which two numeric columns are created based on some data. These two column names are generated by the part of the loop they are in. So for example, the first two columns are df$loop1.1 and df$loop1.2 and then in the following loop they'd be df$loop2.1 and df$loop2.2
I need to create a third column that is based off of those two columns like so:
df[, divisioncol := loop1.1 / loop1.2]
However, I'm getting the error
non-numeric argument to binary operator
Any suggestions?
Solved:
df[, divisioncol := get(loop1.1)/get(loop2.2)]
r data.table
asked Mar 25 at 20:46
JustinJustin
223 bronze badges
I'm running a loop in which two numeric columns are created based on some data. These two column names are generated by the part of the loop they are in. So for example, the first two columns are df$loop1.1 and df$loop1.2 and then in the following loop they'd be df$loop2.1 and df$loop2.2
I need to create a third column that is based off of those two columns like so:
df[, divisioncol := loop1.1 / loop1.2]
However, I'm getting the error
non-numeric argument to binary operator
Any suggestions?
Solved:
df[, divisioncol := get(loop1.1)/get(loop2.2)]
r data.table
r data.table
asked Mar 25 at 20:46
JustinJustin
223 bronze badges
asked Mar 25 at 20:46
JustinJustin
223 bronze badges
edited Mar 25 at 21:00
Justin
asked Mar 25 at 20:46
JustinJustin
223 bronze badges
asked Mar 25 at 20:46
JustinJustin
223 bronze badges
asked Mar 25 at 20:46
JustinJustin
223 bronze badges
223 bronze badges
1
If you are passing the column names, maybe wrap inget, likeget(loop1.1) / get(loop1.2)stackoverflow.com/q/12391950
– Frank
Mar 25 at 20:58
add a comment |
1
If you are passing the column names, maybe wrap inget, likeget(loop1.1) / get(loop1.2)stackoverflow.com/q/12391950
– Frank
Mar 25 at 20:58
1
1
If you are passing the column names, maybe wrap in
get, like get(loop1.1) / get(loop1.2) stackoverflow.com/q/12391950– Frank
Mar 25 at 20:58
If you are passing the column names, maybe wrap in
get, like get(loop1.1) / get(loop1.2) stackoverflow.com/q/12391950– Frank
Mar 25 at 20:58
add a comment |
1 Answer
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Your error is caused by a non-numeric argument to a binary operator. The binary operator here is the function /, which is division. This means that either loop1.1 or loop1.2 contains a non-numeric value. Verify this by typing typeof(df$loop1.1) and typeof(df$loop1.2)
answered Mar 25 at 20:49
Corey LevinsonCorey Levinson
3253 silver badges12 bronze badges
add a comment |
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Your error is caused by a non-numeric argument to a binary operator. The binary operator here is the function /, which is division. This means that either loop1.1 or loop1.2 contains a non-numeric value. Verify this by typing typeof(df$loop1.1) and typeof(df$loop1.2)
answered Mar 25 at 20:49
Corey LevinsonCorey Levinson
3253 silver badges12 bronze badges
add a comment |
Your error is caused by a non-numeric argument to a binary operator. The binary operator here is the function /, which is division. This means that either loop1.1 or loop1.2 contains a non-numeric value. Verify this by typing typeof(df$loop1.1) and typeof(df$loop1.2)
answered Mar 25 at 20:49
Corey LevinsonCorey Levinson
3253 silver badges12 bronze badges
add a comment |
Your error is caused by a non-numeric argument to a binary operator. The binary operator here is the function /, which is division. This means that either loop1.1 or loop1.2 contains a non-numeric value. Verify this by typing typeof(df$loop1.1) and typeof(df$loop1.2)
answered Mar 25 at 20:49
Corey LevinsonCorey Levinson
3253 silver badges12 bronze badges
Your error is caused by a non-numeric argument to a binary operator. The binary operator here is the function /, which is division. This means that either loop1.1 or loop1.2 contains a non-numeric value. Verify this by typing typeof(df$loop1.1) and typeof(df$loop1.2)
answered Mar 25 at 20:49
Corey LevinsonCorey Levinson
3253 silver badges12 bronze badges
answered Mar 25 at 20:49
Corey LevinsonCorey Levinson
3253 silver badges12 bronze badges
answered Mar 25 at 20:49
Corey LevinsonCorey Levinson
3253 silver badges12 bronze badges
answered Mar 25 at 20:49
Corey LevinsonCorey Levinson
3253 silver badges12 bronze badges
3253 silver badges12 bronze badges
add a comment |
add a comment |
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If you are passing the column names, maybe wrap in
get, likeget(loop1.1) / get(loop1.2)stackoverflow.com/q/12391950– Frank
Mar 25 at 20:58