Adaikalavan Ramasamy wrote:
Please give a reproducible example. Here is one way :
# generate example
v1 <- rep( c(0, 2, 10, 11, 13, 14, 15), c(15, 6, 1, 3, 8, 15, 10) ) t1 <- table(v1) t1
v1
0 2 10 11 13 14 15 15 6 1 3 8 15 10
v2 <- rep( c(0, 1, 2, 10, 11, 12, 13, 14, 15), c(817, 119, 524, 96,
700, 66, 559, 358, 283) )
t2 <- table(v2) t2
v2
0 1 2 10 11 12 13 14 15 817 119 524 96 700 66 559 358 283
# find results
merge(t1, t2, by=1, all.x=TRUE)
v1 Freq.x Freq.y 1 0 15 817 2 10 1 96 3 11 3 700 4 13 8 559 5 14 15 358 6 15 10 283 7 2 6 524
Uwe's suggestion may need a slight modification as the two table have different labels/levels and hence non-conformable for division
No!
Uwe's suggestion works perfectly well, since *the same* variable is used for calculating both tables, expect: in one case it's a *subset*. Hence converting to factor at first is completely sufficient!!!
Reproducible example from your v2:
v2 <- rep(c(0, 1, 2, 10, 11, 12, 13, 14, 15),
c(817, 119, 524, 96, 700, 66, 559, 358, 283))
v1 <- factor(v2)
set.seed(1)
v1 <- sample(v1, 10)
table(v1)
table(v2)
table(v1)/table(v2)So, what's the problem???
Uwe Ligges
t2.f <- table( v2.f <- factor(v2) ) t1.f <- table( v1.f <- factor(v1, levels=levels(v2.f)) )
cbind( t1.f, t2.f, ratio=t1.f / t2.f )
t1.f t2.f ratio 0 15 817 0.018359853 1 0 119 0.000000000 2 6 524 0.011450382 10 1 96 0.010416667 11 3 700 0.004285714 12 0 66 0.000000000 13 8 559 0.014311270 14 15 358 0.041899441 15 10 283 0.035335689
Also have a look at this related posting http://tolstoy.newcastle.edu.au/R/help/04/06/0594.html
Regards, Adai.
On Sun, 2004-07-18 at 13:05, Uwe Ligges wrote:
PvR wrote:
Hi all,
I am *completely* lost in trying to solve a relatively simple task.
I want to compute the relative number of occurences of an event, the data of which sits in a large table (read from file).
I have the occurences of the events in a table 'tt'
0 2 10 11 13 14 15 15 6 1 3 8 15 10
.. meaning that event of type '0' occurs 15 times, type '2' occurs 6 times etc.
Now I want to divide the occurence counts by the total number of events of that type, which is given in the table tt2:
0 1 2 10 11 12 13 14 15 817 119 524 96 700 66 559 358 283
Saying that event type '0' occurred 817 times, type '1' occurs 119 times etc.
The obvious problem is that not all events in tt2 are present in tt, which is the result of the experiment so that cannot be changed.
What needs to be done is loop over tt, take the occurence count, and divide that with the corresponding count in tt2. This corresponding tt2 count is *not* at the same index in tt2, so I need a reverse lookup of the type number. For example:
event type 10:
occurs 1 time (from table tt)
occurs 96 times in total (from table tt2) <- this is found by looking up type '10' in tt2 and reading out 96
result: 1/96
I have tried programming this as follows:
It's *much* easier. Just make V32 a factor. After that, table() knows all the levels and counts also the zeros:
V32 <- factor(V32) table(V32[V48 == 0]) / table(V32)
Uwe Ligges
tt <- table(V32[V48 == 0]) # this is taking the events I want counted
tt2 <- table(V32) # this is taking the total event count per type
df <- as.data.frame(tt) #convert to dataframe to allow access to type-numbers .. ?
df2 <- as.data.frame(tt2) #same here
print(tt); print(df);
print(tt2); print(df2);
for( i in 1:length(tt) ) { #loop over smallest table tt
print("i:"); #index
print(i);
print( "denominator "); #corresponds to the "1" in the example
print( df$Freq[i] );
denomtag = ( df$Var1[ i ] ); # corresponds to the "10" in the example, being the type number of the event
print("denomtag ");
print( denomtag );
print( "nominator: " );
print( df2[2][ df[1] == as.numeric(denomtag) ] ); #this fails ....
#result would then be somthing like : denomitor / nominator }
The problem is that the factor names that are extracted in 'denomtag' are not usable as index in the dataframe in the last line. I have tried converting to numeric using 'as.numeric', but that fails since this returns the index in the factor rather then the factor name I need from the list.
Any suggestions .. ? I am sure its dead simple, as always.
Thanks,
Piet (Belgium)
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