Displaying 3 results from an estimated 3 matches for "eqgre".
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egrep
2010 Jan 22
2
column selection in list
Hi everybody!
I have a (stupid) question but I cannot find a way to do it!
I have a list like:
> SPECSHOR_tx_Asfc
$cotau
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...
2010 Jan 21
2
loop on list levels and names
...tau tx 159.081789 18.134533 0.000462
cotau tm 160.641503 6.411332 0.000571
cotau tm 79.238023 3.828254 0.001182
cotau tm 143.20655 11.921899 0.000192
cotau tm 115.476996 33.116386 0.000417
cotau tm 594.256234 72.538131 0.000477
eqgre tx 188.261324 8.279096 0.000777
eqgre tx 152.444216 2.596325 0.001022
eqgre tx 256.601507 8.279096 0.000566
eqgre tx 250.816445 18.134533 0.000535
eqgre tx 272.396711 24.492879 0.000585
eqgre tm 172.63264 4.291884 0.001781
eqgr...
2010 Sep 21
2
labels in (box)plot
...on a boxplot (x and 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,...