Question

我有两个任意人A和B之间的对话记录。

c1 <- "Person A: blabla...something Person B: blabla something else Person A: OK blabla"
c2 <- "Person A: again blabla Person B: blabla something else Person A: thanks blabla"

数据框如下所示：

df <- data.frame(id = rbind(123, 345), conversation = rbind(c1, c2))

df

    id                                                                     conversation
c1 123 Person A: blabla...something Person B: blabla something else Person A: OK blabla
c2 345   Person A: again blabla Person B: blabla something else Person A: thanks blabla

现在我想只提取人A的一部分并将其放在数据框中。结果应该是：

   id                     person_A
1 123 blabla...something OK blabla
2 345   again blabla thanks blabla

Answer 1

我很高兴以某种方式解决这类问题，让您可以访问所有数据（包括人员B的话语）。对于这种列拆分，我喜欢 tidyr 的extract。我过去常常使用do.call(rbind, strsplit()))方法，但我喜欢extract方法的干净程度。

c1 <- "Person A: blabla...something Person B: blabla something else Person A: OK blabla"
c2 <- "Person A: again blabla Person B: blabla something else Person A: thanks blabla"
c3 <- "Person A: again blabla Person B: blabla something else"
df <- data.frame(id = rbind(123, 345, 567), conversation = rbind(c1, c2, c3))


if (!require("pacman")) install.packages("pacman")
pacman::p_load(dplyr, tidyr)

conv <- strsplit(as.character(df[["conversation"]]), "\\s+(?=Person\\s)", perl=TRUE)

df2 <- df[rep(1:nrow(df), sapply(conv, length)), ,drop=FALSE]
rownames(df2) <- NULL
df2[["conversation"]] <- unlist(conv)

df2 %>%
    extract(conversation, c("Person", "Conversation"), "([^:]+):\\s+(.+)")

##    id   Person          Conversation
## 1 123 Person A    blabla...something
## 2 123 Person B blabla something else
## 3 123 Person A             OK blabla
## 4 345 Person A          again blabla
## 5 345 Person B blabla something else
## 6 345 Person A         thanks blabla
## 7 567 Person A          again blabla
## 8 567 Person B blabla something else


df2 %>%
    extract(conversation, c("Person", "Conversation"), "([^:]+):\\s+(.+)") %>%
    filter(Person == "Person A")    

##    id   Person       Conversation
## 1 123 Person A blabla...something
## 2 123 Person A          OK blabla
## 3 345 Person A       again blabla
## 4 345 Person A      thanks blabla
## 5 567 Person A       again blabla

或者在显示所需的输出时折叠它们：

df2 %>%
    extract(conversation, c("Person", "Conversation"), "([^:]+):\\s+(.+)") %>%
    filter(Person == "Person A") %>%
    group_by(id) %>%
    select(-Person) %>%
    summarise(Person_A =paste(Conversation, collapse=" "))

##    id                     Person_A
## 1 123 blabla...something OK blabla
## 2 345   again blabla thanks blabla
## 3 567                 again blabla

编辑：实际上我怀疑你的数据有像＃34; john Smith＆＃34;与＃34;人A＆＃34;。如果是这种情况，这个初始正则表达式拆分将捕获使用大写后跟冒号的名字和姓氏：

c1 <- "Greg Smith: blabla...something Sue Williams: blabla something else Greg Smith: OK blabla"
c2 <- "Greg Smith: again blabla Sue Williams: blabla something else Greg Smith: thanks blabla"
c3 <- "Greg Smith: again blabla Sue Williams: blabla something else"
df <- data.frame(id = rbind(123, 345, 567), conversation = rbind(c1, c2, c3))r


conv <- strsplit(as.character(df[["conversation"]]), "\\s+(?=([A-Z][a-z]+\\s+[A-Z][a-z]+:))", perl=TRUE)

df2 <- df[rep(1:nrow(df), sapply(conv, length)), ,drop=FALSE]
rownames(df2) <- NULL
df2[["conversation"]] <- unlist(conv)

df2 %>%
    extract(conversation, c("Person", "Conversation"), "([^:]+):\\s+(.+)")

##    id       Person          Conversation
## 1 123   Greg Smith    blabla...something
## 2 123 Sue Williams blabla something else
## 3 123   Greg Smith             OK blabla
## 4 345   Greg Smith          again blabla
## 5 345 Sue Williams blabla something else
## 6 345   Greg Smith         thanks blabla
## 7 567   Greg Smith          again blabla
## 8 567 Sue Williams blabla something else

Answer 2

使用stringr包

首先我们使用＆＃34来分割字符串;人物A：＆＃34;作为分隔符

library(stringr)
conv.split <- str_split(df$conversation, "Person A: ")

这将为我们提供由A开始的所有对话，并附上（可选）答案B

我们现在删除B＆B的答案

conv.split <- lapply(conv.split, function(x){str_split(x, "Person B:.*")})

最后我们将每个元素取消列表并将它们一起折叠成一个字符串

sapply(conv.split, function(x){x <- unlist(x); paste(x, collapse = "")})

结果：

[1] "blabla...something OK blabla" "again blabla thanks blabla"

也适用于B开始对话的情况，如果两个人中只有一人说话，也可以进行长时间的对话。

Answer 3

使用基础R的data.table and gsub`：

require(data.table)
setDT(df)[, Person_A := gsub(".*Person A:[ ]*(.*)[ ]*Person B.*:[ ]*(.*)$", 
                         "\\1\\2", conversation)][, conversation := NULL]
df
#     id                       Person_A
# 1: 123 blabla...something OK blabla
# 2: 345   again blabla thanks blabla

Answer 4

它可能不适用于所有情况。尤其是会话从Person B开始。如果是这样，请告诉我。其他尝试

df$person_A <- gsub("Person B.*:|Person A:", "", df$conversation)
df <- data.frame(df$id, df$person_A)

Answer 5

这是我的尝试，我还添加了由人B开始的第二次对话以及由人B结束的对话，以便也涵盖这些情况：

c1 <- "Person A: blabla...something Person B: blabla something else Person A: OK blabla"
c2 <- "Person A: again blabla Person B: blabla something else Person A: thanks blabla"
c3 <- "Person A: again blabla Person B: blabla something else"
df <- data.frame(id = rbind(123, 345, 567), conversation = rbind(c1, c2, c3))


df$PersonA <- gsub("(Person A: |Person B: .+? (?<= Person A: )|Person B: .+?\\Z)", "", df$conversation, perl = TRUE)
df$PersonA

我正在使用gsub删除：

人A：
B组的句子后跟A句
B＆＃39的句子在转换结束时\Z

我使用了perl = TRUE，因为生命太短暂，不能使用后视镜...呃......后视操作员。

如何在两个人A和B之间的对话中仅提取人A的陈述

5 个答案: