[R] Assigning values based on a separate reference (lookup) table

[jim holtman]

You might want to look at 'findInterval'.  This will give you back an
index based on the ranges you specify; e.g., the breaks in you table
that has the 'p' in it.  Here is an example if your table had 100
values between -89.5 & 89.5 and was split evenly:

> x <- seq(-89.5, 89.5, length=50)
> x
 [1] -89.500000 -85.846939 -82.193878 -78.540816 -74.887755 -71.234694
-67.581633 -63.928571 -60.275510 -56.622449 -52.969388
[12] -49.316327 -45.663265 -42.010204 -38.357143 -34.704082 -31.051020
-27.397959 -23.744898 -20.091837 -16.438776 -12.785714
[23]  -9.132653  -5.479592  -1.826531   1.826531   5.479592   9.132653
 12.785714  16.438776  20.091837  23.744898  27.397959
[34]  31.051020  34.704082  38.357143  42.010204  45.663265  49.316327
 52.969388  56.622449  60.275510  63.928571  67.581633
[45]  71.234694  74.887755  78.540816  82.193878  85.846939  89.500000
> z <- runif(100, -89.5, 89.5)
> z
  [1] -41.9739493 -22.8898220  13.0407520  73.0691944 -53.3989343
71.3117536  79.5968731  28.7828049  23.1114139 -78.4402576
 [11] -52.6305511 -57.8963413  33.4770896 -20.7454344  48.3016142
-0.4118357  38.9537130  88.0511910 -21.4737029  49.6626946
 [21]  77.8122364 -51.5264887  27.1496041 -67.0256378 -41.6675003
-20.3855774 -87.1031304 -21.0525557  66.1746614 -28.5775296
 [31]  -3.2076593  17.8222828  -1.1561060 -56.1670493  58.5998240
30.1555461  52.6689351 -70.1780910  40.0442593 -15.8818771
 [41]  57.4493866  26.3237747  50.6449644   9.4934998   5.3198049
51.7947655 -85.3237148  -4.0758184  41.5841592  34.4989486
 [51]  -4.0060876  64.6564964 -11.0806178 -45.6812874 -76.8484506
-71.6955573 -32.8873644   3.3355331  28.9989087 -16.6773965
 [61]  73.9047904 -36.9449963  -7.3272350 -30.0013533  27.0058136
-43.3149963  -3.8404006  47.6696100 -74.4198023  67.1825181
 [71] -28.8059441  60.7598227 -27.4436554 -29.7542874  -4.2331271
70.2035021  65.2167652 -19.6918717  49.6404051  82.4506215
 [81] -11.6959522  38.0401275 -17.9010080 -31.2619648  46.0185995
-53.2180863  37.7906988 -67.7171461 -45.5575560 -63.8485161
 [91] -46.6063347 -78.9507465  25.4695983  67.3521891  49.9257273
53.2182798  -8.0058728 -16.0949493  55.6457735  18.7830590
> findInterval(z, x)
  [1] 14 19 29 45 10 45 47 33 31  4 11  9 34 19 38 25 36 49 19 39 46
11 32  7 14 19  1 19 43 17 24 30 25 10 41 33 39  6 36 21 41 32
 [43] 39 28 26 39  2 24 36 34 24 43 22 12  4  5 16 26 33 20 45 15 23
17 32 13 24 38  5 43 17 42 17 17 24 44 43 20 39 48 22 35 20 16
 [85] 38 10 35  6 13  8 12  3 32 43 39 40 23 21 40 30
>

>From the last statement you have the index into the 'p' and can use
that in the rest of your equation.

On Fri, Aug 14, 2009 at 1:48 PM, Steve Murray<smurray444 at hotmail.com> wrote:
>
> Dear R Users,
>
> I have a data frame of 360 rows by 720 columns (259200 values). For each value in this grid I am hoping to apply an equation to, to generate a new grid. One of the parts of the equation (called 'p') relies on reading from a separate reference table. This is Table 4 at: http://www.fao.org/docrep/s2022e/s2022e07.htm#3.1.3%20blaney%20criddle%20method (scroll down a little).
>
> Therefore, 'p' relies on the latitude of the values in the initial 360 x 720 data frame. The row names of the data frame contain the latitude values and these range from between 89.75 to -89.75 (the latter being South of the Equator).
>
> My question is, how do I go about forming a loop to read each of the 259200 values and assign it a 'p' value (from the associated reference table), based on it's latitude?
>
> My thinking was to do a series of 'if' statements, but this soon got very, very messy - any ideas which get the job done (and aren't a riddle to follow), would be most welcome.
>
> Many thanks for any advice,
>
> Steve
>
>
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>
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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?