3

我有一个要在其上使用表的数据框列表。该列表如下所示：

pronouns <- data.frame(pronounciation = c("juː","juː","juː","ju","ju","jə","jə","hɪm","hɪm","hɪm", "həm","ðɛm"), words = c("you","you","you","you","you","you","you","him","him","him","him","them"))
articles <- data.frame(pronounciation = c("ðiː","ði","ði","ðə","ðə","ði","ðə","eɪ","eɪ","æɪ","æɪ","eɪ","eɪ","eɪ","e"), words = c("the","the","the","the","the","the","the","a","a","a","a","a","a","a","a"))
numbers <- data.frame(pronounciation = c("wʌn","wʌn","wʌn","wʌn","wan","wa:n","tuː","tuː","tuː","tuː","tu","tu","tuː","tuː","θɹiː"), words = c("one","one","one","one","one","one","two","two","two","two","two","two","two","two","three"))
ls <- list(pronouns, articles, numbers)

ls[[1]]
   pronounciation words
1             juː   you
2             juː   you
3             juː   you
4              ju   you
5              ju   you
6              jə   you
7              jə   you
8             hɪm   him
9             hɪm   him
10            hɪm   him
11            həm   him
12            ðɛm  them

从这个数据框列表中，我想使用 table() 提取 $words 的列联表，但还要同时选择每个单词最常见的发音。所需的结果在 ls_out 中：

pronouns_out <- data.frame(pronounciation = c("juː","hɪm","ðɛm"), words = c("you","him","them"), occurence = c(7,4,1))
articles_out <- data.frame(pronounciation = c("ði","eɪ"), words = c("the","a"), occurence = c(7,8))
numbers_out <- data.frame(pronounciation = c("wʌn","tuː","θɹiː"), words = c("one","two","three"), occurence = c(6,8,1))
ls_out <- list(pronouns_out, articles_out, numbers_out)

ls_out[[1]]
  pronounciation words occurence
1            juː   you         7
2            hɪm   him         4
3            ðɛm  them         1

如果两个或多个发音频率相同（如 ls[[2]] 中的 ði 和 ðə），则需要随机选择一个发音。

任何关于如何做到这一点的建议都非常受欢迎。

4

3 回答 3

1

使用table（和lapply）：

ff = function(pronounce, word) 
{
    tab = table(word, pronounce)
    data.frame(pronounciation = colnames(tab)[max.col(tab, "random")], 
               words = rownames(tab),
               occurences = unname(rowSums(tab)))
}

lapply(ls, function(x) ff(x$pronounciation, x$words))

#[[1]]
#     pronounciation words occurences
#1        h<U+026A>m   him          4
#2 <U+00F0><U+025B>m  them          1
#3        ju<U+02D0>   you          7
#
#[[2]]
#  pronounciation words occurences
#1      e<U+026A>     a          8
#2      <U+00F0>i   the          7
#
#[[3]]
#      pronounciation words occurences
#1         w<U+028C>n   one          6
#2 θ<U+0279>i<U+02D0> three          1
#3         tu<U+02D0>   two          8
于 2013-10-24T11:40:41.833 回答
0

使用data.table图书馆 -

library(data.table)

dtlist<-list(pronouns,articles,numbers)
lapply(dtlist,setDT)

# for each data.table in the dtlist, calculate frequency by pron, words
dtlistfreq1 <- 
  lapply(dtlist, function(x) x[,.(freq = .N), by = .(pronunciation,words)])
# for each data.table in the dtlistfreq, pick the highest freq by words
dtlistfreq2 <- 
  lapply(dtlistfreq1, function(x) x[,.SD[which.max(freq)], by = .(words)])

输出

> dtlistfreq2 
[[1]]
   words pronounciation freq
1:   you            ju?    3
2:   him            h?m    4
3:  them            ð?m    1

[[2]]
   words pronounciation freq
1:   the             ði    3
2:     a             e?    5

[[3]]
   words pronounciation freq
1:   one            w?n    4
2:   two            tu?    6
3: three           ??i?    1
于 2013-10-24T11:27:08.083 回答
0

这是一个使用它的解决方案data.table，我认为可以得到你最初想要的东西，其中occurrence每个 的出现总数word，而不是 ( word, pronunciation) 对的数量：

dtlist<-list(pronouns,articles,numbers)
lapply(dtlist,setDT)

common_r<-function(x){
  t<-sort(table(x),decreasing=T)
  n<-length(t[t==max(t)])
  c<-if (n>1)names(t)[ceiling(n*runif(1))] else names(t)[1]
  c
}
lapply(dtlist,function(x)setcolorder(x[,.(occurrence=.N,
                                       pronunciation=common_r(pronunciation)),
                                       by=words]),
                                     c("pronunciation","words","occurrence")))

输出：

[[1]]
   pronunciation words occurrence
1:           juː   you          7
2:           hɪm   him          4
3:           ðɛm  them          1

[[2]]
   pronunciation words occurrence
1:            ði   the          7
2:            eɪ     a          8

[[3]]
   pronunciation words occurrence
1:           wʌn   one          6
2:           tuː   two          8
3:          θɹiː three          1

请注意，当最常见的发音不是唯一的时，我已经注意随机化；如果它总是唯一的（或者如果您不关心在这种情况下选择哪种发音），则可以简化：

common_r<-function(x){names(sort(table(x),decreasing=T))[1]}

如果您不想通过包装为不同的单词类别携带 3 个单独的列表，则可以进一步简化lapply输出rbindlist：

   pronunciation words occurrence
1:           juː   you          7
2:           hɪm   him          4
3:           ðɛm  them          1
4:            ði   the          7
5:            eɪ     a          8
6:           wʌn   one          6
7:           tuː   two          8
8:          θɹiː three          1

我们还可以category在这个新的说法中添加一个字段，data.table每个单词来自哪个单词类别。

于 2015-07-20T21:05:30.180 回答