英文:
Get character indices match in one string and apply to another string
问题
I have the dataframe below, where each row represents changes in text. I then use the adist() function to extract whether the change is a match (M), insertion (I), substitution (S) or deletion (D).
I need to find all of the indices of Is in the change column (illustrated here in the insrtion_idx column). Using those indices, I need to extract the corresponding characters in current_text (illustrated here in insertion_chars).
df <- tibble(current_text = c("A","AB","ABCD","ABZ"),previous_text = c("","A","AB","ABCD"),change = c("I","MI","MMII","MMSD"),insertion_idx = c(c(1),c(2),c(3,4),""),insertion_chars = c("A","B","CD",""))
I have tried splitting up strings and comparing string differences, but this gets very messy very fast with real-world data. How do I accomplish the above task?
英文:
I have the dataframe below, where each row represents changes in text. I then use the adist() function to extract whether the change is a match (M), insertion (I), substitution (S) or deletion (D).
I need to find all of the indices of Is in the change column (illustrated here in the insrtion_idx column). Using those indices, I need to extract the corresponding characters in current_text (illustrated here in insertion_chars).
df <- tibble(current_text = c("A","AB","ABCD","ABZ"),previous_text = c("","A","AB","ABCD"),change = c("I","MI","MMII","MMSD"),insertion_idx = c(c(1),c(2),c(3,4),""),insertion_chars = c("A","B","CD",""))
I have tried splitting up strings and comparing string differences, but this gets very messy very fast with real-world data. How do I accomplish the above task?
答案1
得分: 3
将我关于使用 gregexpr 和 regmatches 的评论转化为答案。
这个过程的很多部分与这个问题的内容非常相似 - https://stackoverflow.com/questions/2192316/extract-a-regular-expression-match/23901600 - 如果你正在寻找替代方法。
df <- data.frame(current_text = c("A","AB","ABCD","ABZ"),previous_text = c("","A","AB","ABCD"),change = c("I","MI","MMII","MMSD"))df$insertion_idx <- gregexpr("I", df$change)df$insertion_chars <- sapply(regmatches(df$current_text, df$insertion_idx),paste, collapse="")df## current_text previous_text change insertion_chars insertion_idx##1 A I A 1##2 AB A MI B 2##3 ABCD AB MMII CD 3, 4##4 ABZ ABCD MMSD -1
英文:
Turning my comment about using gregexpr and regmatches into an answer.
A lot of this procedure is very similar to the content in this question - https://stackoverflow.com/questions/2192316/extract-a-regular-expression-match/23901600 - if you are looking for alternative methods.
df <- data.frame(current_text = c("A","AB","ABCD","ABZ"),previous_text = c("","A","AB","ABCD"),change = c("I","MI","MMII","MMSD"))df$insertion_idx <- gregexpr("I", df$change)df$insertion_chars <- sapply(regmatches(df$current_text, df$insertion_idx),paste, collapse="")df## current_text previous_text change insertion_chars insertion_idx##1 A I A 1##2 AB A MI B 2##3 ABCD AB MMII CD 3, 4##4 ABZ ABCD MMSD -1
答案2
得分: 2
尝试这个替代方案，以取代thelatemail（很棒的）的建议（也适用）：
quux <- structure(list(current_text = c("A", "AB", "ABCD", "ABZ"), previous_text = c("", "A", "AB", "ABCD"), change = c("I", "MI", "MMII", "MMSD")), row.names = c(NA, -4L), class = c("tbl_df", "tbl", "data.frame"))quux$insertion_idx <- lapply(strsplit(quux$change, ""), function(z) which(z == "I"))quux$insertion_chars <- mapply(function(ctxt, idx) {if (length(idx)) paste(substring(ctxt, idx, idx), collapse = "") else ""}, quux$current_text, quux$insertion_idx)quux# # A tibble: 4 × 5# current_text previous_text change insertion_idx insertion_chars# <chr> <chr> <chr> <list> <chr># 1 A "" I <int [1]> "A"# 2 AB "A" MI <int [1]> "B"# 3 ABCD "AB" MMII <int [2]> "CD"# 4 ABZ "ABCD" MMSD <int [0]> ""
注意，insertion_idx 是一个包含你寻找的索引的列表列：
str(quux)# tibble [4 × 5] (S3: tbl_df/tbl/data.frame)# $ current_text : chr [1:4] "A" "AB" "ABCD" "ABZ"# $ previous_text : chr [1:4] "" "A" "AB" "ABCD"# $ change : chr [1:4] "I" "MI" "MMII" "MMSD"# $ insertion_idx :List of 4# ..$ : int 1# ..$ : int 2# ..$ : int [1:2] 3 4# ..$ : int(0)# $ insertion_chars: Named chr [1:4] "A" "B" "CD" ""# ..- attr(*, "names")= chr [1:4] "A" "AB" "ABCD" "ABZ"
英文:
Try this alternative to thelatemail's (excellent) recommendation (which also works):
quux <- structure(list(current_text = c("A", "AB", "ABCD", "ABZ"), previous_text = c("", "A", "AB", "ABCD"), change = c("I", "MI", "MMII", "MMSD")), row.names = c(NA, -4L), class = c("tbl_df", "tbl", "data.frame"))quux$insertion_idx <- lapply(strsplit(quux$change, ""), function(z) which(z == "I"))quux$insertion_chars <- mapply(function(ctxt, idx) {if (length(idx)) paste(substring(ctxt, idx, idx), collapse = "") else ""}, quux$current_text, quux$insertion_idx)quux# # A tibble: 4 × 5# current_text previous_text change insertion_idx insertion_chars# <chr> <chr> <chr> <list> <chr># 1 A "" I <int [1]> "A"# 2 AB "A" MI <int [1]> "B"# 3 ABCD "AB" MMII <int [2]> "CD"# 4 ABZ "ABCD" MMSD <int [0]> ""
Note that insertion_idx is a list-column with the indices you were looking for:
str(quux)# tibble [4 × 5] (S3: tbl_df/tbl/data.frame)# $ current_text : chr [1:4] "A" "AB" "ABCD" "ABZ"# $ previous_text : chr [1:4] "" "A" "AB" "ABCD"# $ change : chr [1:4] "I" "MI" "MMII" "MMSD"# $ insertion_idx :List of 4# ..$ : int 1# ..$ : int 2# ..$ : int [1:2] 3 4# ..$ : int(0)# $ insertion_chars: Named chr [1:4] "A" "B" "CD" ""# ..- attr(*, "names")= chr [1:4] "A" "AB" "ABCD" "ABZ"
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