Hi:

I think the idea here is to use the cdf intervals as the lookup table for
rand (the entire
vector) and then return the seq value corresponding to the found interval.
Combining
the ideas from Jim Holtman (use findInterval()) and Josh Wiley, what worked
for me
was to split the OP's data into three vectors. Starting from Josh's
datasamp,

datasamp <- structure(list(cdf = c(0, 0.00156, 0.0155, 0.053, 0.116, 0.197,
0.288, 0.38, 0.468, 0.548, 0.618, 0.679, 0.732, 0.776, 0.813,
0.844, 0.87, 0.892, 0.91, 0.926, 0.938, 0.949, 0.957, 0.965,
0.971, 0.976, 0.98, 0.983, 0.986, 0.988, 0.99), seq = c(0L, 20L,
40L, 60L, 80L, 100L, 120L, 140L, 160L, 180L, 200L, 220L, 240L,
260L, 280L, 300L, 320L, 340L, 360L, 380L, 400L, 420L, 440L, 460L,
480L, 500L, 520L, 540L, 560L, 580L, 600L), rand = c(0.262123478,
0.964293344, 0.494827113, 0.733726005, 0.800408948, 0.925748466,
0.047578356, 0.266060366, 0.125522629, 0.701193274, 0.915799432,
0.735984375, 0.517838069, 0.042085025, 0.568346202, 0.067140721,
0.71437727, 0.712210456, 0.288460952, 0.185857723, 0.108859523,
0.513351848, 0.22214423, 0.500350963, 0.437818537, 0.851771599,
0.803521836, 0.249824519, 0.859248634, 0.126926481, 0.713431196
)), .Names = c("cdf", "seq", "rand"), class = "data.frame", row.names =
c(NA,
-31L))

cdf <- datasamp[, 1]
sq <- datasamp[, 2]
rand <- c(datasamp$rand, 0.253425703, 0.830195013, 0.723929563,
            0.027588733, 0.091067232, 0.689504685, 0.890469069, 0.710440382)
rm(datasamp)
cbind(rand, sq[findInterval(rand, cdf)], sq[findInterval(rand, cdf) + 1])

            rand
 [1,] 0.26212348 100 120
 [2,] 0.96429334 440 460
 [3,] 0.49482711 160 180
 [4,] 0.73372600 240 260
 [5,] 0.80040895 260 280
 [6,] 0.92574847 360 380
 [7,] 0.04757836  40  60
...
I noticed that the original result seemed to be an interval below what the
OP
expected, so add one to the vector of indices returned from findInterval()
to
get the third column.

HTH,
Dennis



On Thu, May 27, 2010 at 6:20 PM, Roslina Zakaria <zroslina@yahoo.com> wrote:

> Hi R-users,
>
> I would like to search for the values of seq that match my rand values.  In
> excel I will use =VLOOKUP(G2,$E$2:$F$32,2).  For example, for rand=.262 it
> will give me approximately seq=120 and rand=0.964293344, seq=460 and etc.
>
> E           F     G
> cdf         seq   rand
> 0.00E+00    0     0.262123478
> 1.56E-03    20    0.964293344
> 1.55E-02    40    0.494827113
> 5.30E-02    60    0.733726005
> 1.16E-01    80    0.800408948
> 1.97E-01    100   0.925748466
> 2.88E-01    120   0.047578356
> 3.80E-01    140   0.266060366
> 4.68E-01    160   0.125522629
> 5.48E-01    180   0.701193274
> 6.18E-01    200   0.915799432
> 6.79E-01    220   0.735984375
> 7.32E-01    240   0.517838069
> 7.76E-01    260   0.042085025
> 8.13E-01    280   0.568346202
> 8.44E-01    300   0.067140721
> 8.70E-01    320   0.71437727
> 8.92E-01    340   0.712210456
> 9.10E-01    360   0.288460952
> 9.26E-01    380   0.185857723
> 9.38E-01    400   0.108859523
> 9.49E-01    420   0.513351848
> 9.57E-01    440   0.22214423
> 9.65E-01    460   0.500350963
> 9.71E-01    480   0.437818537
> 9.76E-01    500   0.851771599
> 9.80E-01    520   0.803521836
> 9.83E-01    540   0.249824519
> 9.86E-01    560   0.859248634
> 9.88E-01    580   0.126926481
> 9.90E-01    600   0.713431196
>                   0.253425703
>                   0.830195013
>                   0.723929563
>                   0.027588733
>                   0.091067232
>                   0.689504685
>                   0.890469069
>                   0.710440382
> Thank you so much for your help.
>
>
>
>
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>
>
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> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>

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