我不太清楚该怎么回答我的问题。我想要做的是创建一个循环,将数据帧行中的每个值与另一个数据帧中的键匹配,并将该行的每一列中的键值相加,并将其存储在一个新的数据帧中,该数据帧的大小与键的大小相同。
用一个例子来解释应该容易得多。我是一个完全新手的R和编程,并仍在学习词汇表。
我有一个数据的词,其中每列对应于一个音素(独特的语音)。
Words_DF <- data.frame( word = c("CAT", "BAT", "APPLE"), Phoneme1 = c("K", "B", "AE"), Phoneme2 = c("AE", "AE", "P"), Phoneme3 = c("T", "T", "AH"), Phoneme4 = c("Null", "Null", "L"))
word Phoneme1 Phoneme2 Phoneme3 Phoneme4
1 CAT K AE T Null
2 BAT B AE T Null
3 APPLE AE P AH L我有另一个数据框架,其中每个音素对应一系列二进制值。
Phoneme_DF <- data.frame( phoneme = c("AE", "AH", "B", "K", "T", "P", "L"), is_consonant = c(0, 0, 1, 1, 1, 1, 1), is_labial = c(0, 0, 0, 0, 0, 1, 0))
phoneme is_consonant is_labial
1 AE 0 0
2 AH 0 0
3 B 1 1
4 K 1 0
5 T 1 0
6 P 1 1
7 L 1 0我正试图找出一条贯穿我的Words_DF每一行的方法,并在我的Phoneme_DF中的每个音素列中查找值,并将它们相加到一个新的数据框架中,该数据框架如下所示:
New_DF <- data.frame( word = c("CAT", "BAT", "APPLE"), consonants_in_word = c(2, 2, 3), labials_in_word = c(0, 1, 1))
word consonants_in_word labials_in_word
1 CAT 2 0
2 BAT 2 1
3 APPLE 2 1我尝试过编写某种循环,遍历Words_DF的每一行,在每一行中遍历每一列,并在Phoneme_DF中查找该值,然后进行和。
New_DF <- data.frame( word = c("CAT", "BAT", "APPLE"), consonants_in_word = c(0, 0 , 0 ), labials_in_word = c(0, 0, 0))
for(i in 1:length(SAMPLE_Words)){
for(j in 1:length(where(SAMPE_Words[[j]]) %in% SAMPLE_Phoneme_DF[i])) {
rbind(New_DF, sum(Phoneme_DF[i, ]))
}
}我希望我的问题是有道理的。谢谢你的帮助!)
发布于 2017-11-20 01:27:05
我想你的输出是关的,Apple应该只有两个辅音。试试这个:
library(tidyverse)
Words_DF %>%
gather(value, key, -word) %>%
left_join(Phoneme_DF, by = c("key" = "phoneme")) %>%
group_by(word) %>%
mutate(consonants_in_word = sum(is_consonant, na.rm = TRUE),
labials_in_word = sum(is_labial, na.rm = TRUE)) %>%
distinct(word, .keep_all = TRUE) %>%
select(word, consonants_in_word, labials_in_word)返回:
# A tibble: 3 x 3
# Groups: word [3]
word consonants_in_word labials_in_word
<chr> <int> <int>
1 CAT 2 0
2 BAT 2 1
3 APPLE 2 1这是我使用的数据:
Words_DF <- read.table(text = "word Phoneme1 Phoneme2 Phoneme3 Phoneme4
1 CAT K AE T Null
2 BAT B AE T Null
3 APPLE AE P AH L",
stringsAsFactors = FALSE, header = TRUE)
Phoneme_DF <- read.table(text = "phoneme is_consonant is_labial
1 AE 0 0
2 AH 0 0
3 B 1 1
4 K 1 0
5 T 1 0
6 P 1 1
7 L 1 0",
stringsAsFactors = FALSE, header = TRUE)发布于 2017-11-20 09:47:42
我有data.table对应的,对于任何感兴趣的人:
Phoneme_DF[melt(Words_DF,id.vars = "word", value.name = "phoneme"), on = "phoneme"][
,lapply(.SD,function(x){sum(x,na.rm = TRUE)}),
.SDcols = c("is_consonant","is_labial"),by = word]给出
word is_consonant is_labial
1: CAT 2 0
2: BAT 2 1
3: APPLE 2 1过程类似于tyluRp提出的方法:以长格式重新构造wordDF数据表,将其与另一个数据表连接,然后逐字对辅音和下唇的值进行求和。
https://stackoverflow.com/questions/47383655
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