自昨晚以来，我一直忙于这个问题，我无法弄清楚如何去做。
我想要做的是将df1字符串与df2字符串匹配并获得类似的字符串
我所做的就像这样
# a function to arrange the data to have IDs for each string
normalize <- function(x, delim) {
x <- gsub(")", "", x, fixed=TRUE)
x <- gsub("(", "", x, fixed=TRUE)
idx <- rep(seq_len(length(x)), times=nchar(gsub(sprintf("[^%s]",delim), "", as.character(x)))+1)
names <- unlist(strsplit(as.character(x), delim))
return(setNames(idx, names))
}
# a function to arrange the second df
lookup <- normalize(df2[,1], ",")
# a function to match them and give the IDs
process <- function(s) {
lookup_try <- lookup[names(s)]
found <- which(!is.na(lookup_try))
pos <- lookup_try[names(s)[found]]
return(paste(s[found], pos, sep="-"))
#change the last line to "return(as.character(pos))" to get only the result as in the comment
}
然后我得到这样的结果
res <- lapply(colnames(df1), function(x) process(normalize(df1[,x], ";")))
这给出了df1中每个字符串的行号和匹配的df2的字符串的行号。所以这个数据的输出看起来像这样
> res
$s1
[1] "3-4" "4-1" "5-4"
$s2
[1] "2-4" "3-15" "7-16"
第一列 ID 是与df1中的字符串匹配的df2的行号 第二列否是匹配的次数 第三列 ID-col-n 是df1中与该字符串匹配的字符串的行号+其列名 第四个是df1第一列的字符串，与该字符串匹配 第五列是与该字符串匹配的第二列的字符串 等等
答案 0 :(得分：6)
在这种情况下，我发现将数据切换为宽格式并将其合并到查找表之前更容易。
你可以尝试：
library(tidyr)
library(dplyr)
df1_tmp <- df1
df2_tmp <- df2
#add numerical id to df1_tmp to keep row information
df1_tmp$id <- seq_along(df1_tmp[,1])
#switch to wide and unnest rows with several strings
df1_tmp <- gather(df1_tmp,key="s_val",value="query_string",-id)
df1_tmp <- df1_tmp %>%
mutate(query_string = strsplit(as.character(query_string), ";")) %>%
unnest(query_string)
df2_tmp$IDs. <- gsub("[()]", "", df2_tmp$IDs.)
#add numerical id to df1_tmp to keep row information
df2_tmp$id <- seq_along(df2_tmp$IDs.)
#unnest rows with several strings
df2_tmp <- df2_tmp %>%
mutate(IDs. = strsplit(as.character(IDs.), ",")) %>%
unnest(IDs.)
res <- merge(df1_tmp,df2_tmp,by.x="query_string",by.y="IDs.")
res$ID_col_n <- paste(paste0(res$id.x,res$s_val))
res$total_id <- 1:nrow(res)
res <- spread(res,s_val,value=query_string,fill=NA)
res
#summarize to get required output
res <- res %>% group_by(id.y) %>%
mutate(No=n()) %>% group_by(id.y,No) %>%
summarise_each(funs(paste(.[!is.na(.)],collapse=","))) %>%
select(-id.x,-total_id)
colnames(res)[colnames(res)=="id.y"]<-"IDs"
res$df1_colMatch_counts <- rowSums(res[,-(1:3)]!="")
df2_counts <- df2_tmp %>% group_by(id) %>% summarize(df2_string_counts=n())
res <- merge(res,df2_counts,by.x="IDs",by.y="id")
res
res
IDs No ID_col_n s1 s2 df1_colMatch_counts df2_string_counts
1 1 1 4s1 P41182 1 2
2 2 1 4s1 P41182 1 2
3 3 1 4s1 P41182 1 2
4 4 3 2s2,3s1,5s1 Q9Y6Q9,Q09472 Q92831 2 4
5 15 1 3s2 P54612 1 5
6 16 1 7s2 O15143 1 7
答案 1 :(得分：2)
对于这个解决方案，我使用一个通用的帮助函数来解决rbind()的限制，它不能处理不一致的列名。您可以使用Combine two data frames by rows (rbind) when they have different sets of columns中的任何函数，但我也写了自己的函数：
rbind.cn <- function(...,filler=NA) {
## note: must explicitly set proper S3 class; otherwise, filler would corrupt the column type in cases where a column is missing from the first df
## for example, logical NA (the default for filler) would nix factors, resulting in character after type promotion
## to do this, will use single-row temp df as first argument to rbind()
## note: tried zero-row to prevent need to excise afterward, but zero-row rbind() arguments are ignored for typing purposes
l <- list(...);
schema <- do.call(cbind,unname(lapply(l,function(df) df[1L,,drop=F]))); ## unname() is necessary, otherwise cbind() tries to be a good citizen and concats first df cell value found in lapply() names onto schema column names
schema <- schema[unique(names(schema))];
res <- do.call(rbind,c(list(schema),lapply(l,function(df) {
cns.add <- names(schema)[!names(schema)%in%names(df)];
do.call(cbind,c(
list(df),
setNames(rep(filler,length(cns.add)),cns.add),
stringsAsFactors=F
));
})))[-1L,,drop=F];
## fix up row names
rns <- do.call(c,lapply(l,rownames));
rownames(res) <- ifelse(grepl('^[0-9]+$',rns),seq_along(rns),rns);
res;
};
我还编写了自己的规范化函数。我认为您通过预先计算两个输入data.frames的规范化表示来处于正确的轨道，但是因为您使用命名向量索引来匹配ID，所以您没有在df2中捕获重复名称的情况，这就是为什么你的结果缺少额外出现的id P41182。这是我的规范化功能：
## normalization function
## for each column, splits on sep and captures the id, row index, column index, and column name in a data.frame
normalize <- function(df,sep) {
do.call(rbind,lapply(seq_along(df),function(ci) {
l <- strsplit(gsub('[()]','',df[[ci]]),sep);
cbind(
do.call(rbind,lapply(seq_along(l),function(ri)
if (length(l[[ri]]) > 0L)
data.frame(id=l[[ri]],ri=ri,stringsAsFactors=F)
)),
ci,
cn=names(df)[ci]
);
}));
};
这是完整的解决方案：
## normalize both data.frames
df1.norm <- normalize(df1,';');
df2.norm <- normalize(df2,',');
## join them on matching ids
df.match <- merge(df1.norm,df2.norm,'id',suffixes=c('.1','.2'));
df.match <- df.match[with(df.match,order(ri.2,cn.1,ri.1)),]; ## order by df2 row index, df1 column name, and finally df1 row index, as per required output
## aggregate and format as required
res <- do.call(rbind.cn,c(by(df.match,df.match$ri.2,function(x) {
strCols <- aggregate(id~cn.1,x[c('id','cn.1')],paste,collapse=','); ## conveniently, automatically orders by the grouping column cn.1
do.call(cbind,c(
list(data.frame(IDs=x$ri.2[1L],No=nrow(x),`ID-col-n`=paste0(x$ri.1,x$cn.1,collapse=','),stringsAsFactors=F,check.names=F)),
setNames(strCols$id,paste0('string-df1-',strCols$cn.1)),
stringsAsFactors=F
));
}),filler='-'));
## order string-df1 columns
res <- res[c(1:3,order(as.integer(sub('.*?([0-9]+)$','\\1',names(res)[-1:-3])))+3L)];
以下是所有中间和最终的data.frames：
df1.norm;
## id ri ci cn
## 1 Q9Y6W5 1 1 s1
## 2 Q9Y6U3 2 1 s1
## 3 Q9Y6Q9 3 1 s1
## 4 P41182 4 1 s1
## 5 Q9HCP0 4 1 s1
## 6 Q09472 5 1 s1
## 7 Q9Y6I3 6 1 s1
## 8 Q9Y6H1 7 1 s1
## 9 Q5T1J5 7 1 s1
## 10 Q16835 1 2 s2
## 11 P61809 2 2 s2
## 12 Q92831 2 2 s2
## 13 P41356 3 2 s2
## 14 P54612 3 2 s2
## 15 A41PH2 3 2 s2
## 16 P3R117 4 2 s2
## 17 P31908 5 2 s2
## 18 P54112 6 2 s2
## 19 O15143 7 2 s2
df2.norm;
## id ri ci cn
## 1 P41182 1 1 IDs.
## 2 P56524 1 1 IDs.
## 3 P41182 2 1 IDs.
## 4 Q9UQL6 2 1 IDs.
## 5 P41182 3 1 IDs.
## 6 Q8WUI4 3 1 IDs.
## 7 Q92793 4 1 IDs.
## 8 Q09472 4 1 IDs.
## 9 Q9Y6Q9 4 1 IDs.
## 10 Q92831 4 1 IDs.
## 11 P30561 5 1 IDs.
## 12 P53762 5 1 IDs.
## 13 Q15021 6 1 IDs.
## 14 Q9BPX3 6 1 IDs.
## 15 Q15003 6 1 IDs.
## 16 O95347 6 1 IDs.
## 17 Q9NTJ3 6 1 IDs.
## 18 Q92902 7 1 IDs.
## 19 Q9NQG7 7 1 IDs.
## 20 Q969F9 8 1 IDs.
## 21 Q9UPZ3 8 1 IDs.
## 22 Q86YV9 8 1 IDs.
## 23 Q92903 9 1 IDs.
## 24 Q96NY9 9 1 IDs.
## 25 Q91VB4 10 1 IDs.
## 26 P59438 10 1 IDs.
## 27 Q8BLY7 10 1 IDs.
## 28 Q92828 11 1 IDs.
## 29 Q13227 11 1 IDs.
## 30 O15379 11 1 IDs.
## 31 O75376 11 1 IDs.
## 32 O60907 11 1 IDs.
## 33 Q9BZK7 11 1 IDs.
## 34 P78537 12 1 IDs.
## 35 Q6QNY1 12 1 IDs.
## 36 Q6QNY0 12 1 IDs.
## 37 Q9NUP1 12 1 IDs.
## 38 Q96EV8 12 1 IDs.
## 39 Q8TDH9 12 1 IDs.
## 40 Q9UL45 12 1 IDs.
## 41 O95295 12 1 IDs.
## 42 O55102 13 1 IDs.
## 43 Q9CWG9 13 1 IDs.
## 44 Q5U5M8 13 1 IDs.
## 45 Q8VED2 13 1 IDs.
## 46 Q91WZ8 13 1 IDs.
## 47 Q8R015 13 1 IDs.
## 48 Q9R0C0 13 1 IDs.
## 49 Q9Z266 13 1 IDs.
## 50 P30561 14 1 IDs.
## 51 O08915 14 1 IDs.
## 52 P07901 14 1 IDs.
## 53 P11499 14 1 IDs.
## 54 Q8WMR7 15 1 IDs.
## 55 P67776 15 1 IDs.
## 56 P11493 15 1 IDs.
## 57 P54612 15 1 IDs.
## 58 P54613 15 1 IDs.
## 59 P61160 16 1 IDs.
## 60 P61158 16 1 IDs.
## 61 O15143 16 1 IDs.
## 62 O15144 16 1 IDs.
## 63 O15145 16 1 IDs.
## 64 P59998 16 1 IDs.
## 65 O15511 16 1 IDs.
df.match;
## id ri.1 ci.1 cn.1 ri.2 ci.2 cn.2
## 2 P41182 4 1 s1 1 1 IDs.
## 4 P41182 4 1 s1 2 1 IDs.
## 3 P41182 4 1 s1 3 1 IDs.
## 8 Q9Y6Q9 3 1 s1 4 1 IDs.
## 6 Q09472 5 1 s1 4 1 IDs.
## 7 Q92831 2 2 s2 4 1 IDs.
## 5 P54612 3 2 s2 15 1 IDs.
## 1 O15143 7 2 s2 16 1 IDs.
res;
## IDs No ID-col-n string-df1-s1 string-df1-s2
## 1 1 1 4s1 P41182 -
## 2 2 1 4s1 P41182 -
## 3 3 1 4s1 P41182 -
## 4 4 3 3s1,5s1,2s2 Q9Y6Q9,Q09472 Q92831
## 5 15 1 3s2 - P54612
## 6 16 1 7s2 - O15143