我已阅读FAQ但现在仍然清楚使用密钥与在合理大数据的连续列表中使用该密钥有什么影响。表？
从我的实验中，我只看到了表现，但不确定是否还有其他事情。
# install.packages(c("data.table", "stringi"), dependencies = TRUE)
library(data.table)
library(stringi)
download.file("https://www.ssa.gov/oact/babynames/state/namesbystate.zip", dest="namesbystate.zip", mode="wb")
unzip("namesbystate.zip", exdir=".")
# Read the list of all text files in variable "filelist"
filelist = list.files(path=".",pattern = ".*.TXT")
colnamelist=c("State","gender","year","name","frequency")
#Read the CSV from all the text files into a data.frame
babynames =lapply(filelist, FUN=read.csv, header=FALSE,col.name=colnamelist);
nametable = rbindlist(babynames,use.names = FALSE,fill = FALSE)
DT = data.table(nametable)
dim(DT) #[1] 5647426 5
setkey(DT,NULL)
system.time(head(DT[,( stri_length(name)),by=c("name", "year")]))
# user system elapsed
# 156.47 0.03 157.64
setkey(DT,year)
system.time(head(DT[,( stri_length(name)),by=name]))
# user system elapsed
# 8.90 0.00 8.99
两种情况下的输出相同
name year V1
1: Anna 1910 4
2: Annie 1910 5
3: Dorothy 1910 7
4: Elsie 1910 5
5: Helen 1910 5
6: Lucy 1910 4