Displaying 3 results from an estimated 3 matches for "gicam".
Did you mean:
giam
2010 Jan 22
2
column selection in list
...SPECSHOR Asfc.median
38 cotau 381.0247
39 cotau 154.6280
40 cotau 303.3219
41 cotau 351.2933
42 cotau 156.5327
$eqgre
SPECSHOR Asfc.median
145 eqgre 219.5389
146 eqgre 162.5926
147 eqgre 146.3726
148 eqgre 127.6413
149 eqgre 274.2888
$gicam
SPECSHOR Asfc.median
263 gicam 174.7445
264 gicam 83.4821
265 gicam 157.6005
266 gicam 153.7519
267 gicam 344.9775
I would just like to remove the column "SPECSHOR" (or extract the other
one) so that it looks like
$cotau
Asfc.median
38 381...
2010 Jan 21
2
loop on list levels and names
...gre tx 272.396711 24.492879 0.000585
eqgre tm 172.63264 4.291884 0.001781
eqgre tm 189.441097 14.425498 0.001347
eqgre tm 170.743788 13.564472 0.000602
eqgre tm 158.960849 10.385299 0.001189
eqgre tm 80.972408 3.828254 0.000644
gicam tx 294.494001 9.656738 0.000524
gicam tx 267.126765 19.128024 0.000647
gicam tx 81.888658 4.782006 0.000492
gicam tx 168.32908 12.729939 0.001097
gicam tx 123.296056 7.007427 0.000659
gicam tm 94.264887 18.134533 0.000752
gicam...
2010 Sep 21
2
labels in (box)plot
...y) to be labeled
I have checked all the par() arguments but couldn't find what I'm
looking for
Here is an example to show it:
df <- structure(list(SPECSHOR = structure(c(1L, 1L, 1L, 3L, 3L, 3L, 3L,
3L, 4L, 4L), .Label = c("cotau", "dibic", "eqgre", "gicam"), class =
"factor"), Sq122.median = c(2.335835, 1.76091, 1.64717, 1.285505,
1.572405, 1.86761, 1.82541, 1.62458, 0.157813, 0.864523)), .Names =
c("SPECSHOR", "Sq122.median"), class = "data.frame", row.names = c(9L,
16L, 23L, 74L, 83L, 90L, 98L, 109L...