On Nov 30, 2010, at 11:40 AM, Graves, Gregory wrote:
Having given myself carpal tunnel looking for answer to this ...
I have a dataset
Named.... what?
each column of which has 12 rows in it. I created a
variable 'z' as follows:
z=1:24 s
Why did you put twice as many elements in z as there are in a column?
Since I have a large number of these plots to make, and they are a bit
complex, I want to want to reference the column I want to plot via a
variable containing the name of that column. As follows:
similar='1978'
s=paste('Y',similar,sep='')
variable s now contains 'Y1978' which is the name of one of the
columns.
However, when I try to plot
plot(z,s,type='l')
Try this ... assuming that there really are 12 item length columns in
a dataframe named, dfm.
plot(1:12, dfm[[s]], type="l")
dataframes are lists that can be accessed by the names of columns
which are interpreted. Don't assume that you can get such
interpretation with the $ operator.
I get a 'x and y lengths differ' error because variable s is being
recognized as 'Y1978' length=1, rather than the contents of the column
Y1978 length=12.
I tried all the usual tricks I know like &s.
Huh? "&" is a logical operator.
How do you get R to
reference a variable as a column name?
Thank you.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
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