我的R数据框有数百行
word Freq
seed 4
seeds 3
contract 2
contracting 2
river 1
我想按照模式对数据进行分组,比如种子+种子......看起来像
word Freq
seed 7
contract 4
river 1
答案 0 :(得分:3)
这可能是另一种方式。在SnowballC
包中,有一个函数可以清理单词并获取单词词干(即wordStem()
)。使用它,我认为你可以跳过字符串操作。完成此过程后,您所要做的就是获得字频的总和。
library(SnowballC)
library(dplyr)
mydf <- read.table(text = "word Freq
seed 4
seeds 3
contract 2
contracting 2
river 1", header = T)
mutate(mydf, word = wordStem(word)) %>%
group_by(word) %>%
summarise(total = sum(Freq))
# word total
# (chr) (int)
#1 contract 4
#2 river 1
#3 seed 7
答案 1 :(得分:1)
一种选择是创建一个分组变量&#39; gr&#39;通过根据&#39; word&#39;中的最小字符数提取子字符串,再用#39; word&#39; sp我们可以得到每组单词的子字符串,然后得到sum
的&#39; Freq&#39;通过&#39; word&#39;。
library(dplyr)
df1 %>%
group_by(gr= substr(word, 1, min(nchar(word)))) %>%
group_by(word= substr(word, 1, min(nchar(word)))) %>%
summarise(Freq= sum(Freq))
word Freq
# (chr) (int)
#1 contract 4
#2 river 1
#3 seed 7
答案 2 :(得分:1)
也可以使用交叉连接,这比上述方法更安全。
2015-10-23T03:34:40
答案 3 :(得分:1)
尝试使用adist
来匹配这些条款。
dat$grp <- seq(nrow(dat))
# generate a matrix comparing the vector of words to themselves
tmp <- adist(dat$word, dat$word, partial=TRUE)
diag(tmp) <- Inf
dat$grp[col(tmp)[tmp==0]] <- row(tmp)[tmp==0]
final <- aggregate(Freq ~ grp, data=dat, sum)
final$word <- dat$word[match(final$grp, dat$grp)]
# grp Freq word
#1 1 7 seed
#2 3 4 contract
#3 5 1 river
使用的数据:
dat <- data.frame(word=c("seed","seeds","contract","contracting","river"),Freq=c(4,3,2,2,1))