Vector subsetting performance: name versus index

Question

If I have a vector v with names:

John       Murray     Lisa       Mike       Joe       Ann 
0.0832090  0.0475580 -0.2797860  0.1086225  0.0104590 -0.0028250 

What is time complexity of v['Joe'] versus v[4]? I guess the former would take O(log n) as it should involves binary search, but I'm still not sure whether the latter is O(1) or not.

Also, does the result generalize to the case when v is a list/data frame rather than an atomic vector?

Answer 1

It seems to be approximately O(n), i.e. a vector scan, in the case of name look ups. Your conjecture of O(1) for lookup using indices seems sound...

#  Unique names for longish vector
nms <- apply( expand.grid( letters , letters , letters , letters ) , 1 , paste , collapse = "" )
length(nms)
#[1] 456976
length(unique(nms))
#[1] 456976

#  Start of names
head(nms)
#[1] "aaaa" "baaa" "caaa" "daaa" "eaaa" "faaa"

#  End of names
tail(nms)
#[1] "uzzz" "vzzz" "wzzz" "xzzz" "yzzz" "zzzz"

#  Large named vector
x <- setNames( runif( 456976 ) , nms )

#  Small named vector
y <- setNames( runif(26) , letters )

#  Timing information
require( microbenchmark )
bm <- microbenchmark( x['daaa'] , x[4] , x['vzzz'] , x[456972] , y['d'] , y[4] )
print( bm , order = 'median' , unit = 'relative' , digits = 3 )
#Unit: relative
#      expr min       lq   median       uq      max neval
# x[456972] NaN 1.00e+00     1.00     1.00    1.000   100
#      x[4] Inf 1.00e+00     1.33     1.07    0.957   100
#      y[4] NaN 5.01e-01     1.33     1.14    0.191   100
#    y["d"] Inf 1.00e+00     2.00     1.25    0.265   100
# x["vzzz"] Inf 6.57e+04 44412.24  9969.64 3439.154   100
# x["daaa"] Inf 6.59e+04 44582.73 10049.63 1207.337   100