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我想删除彼此之后具有相同两个或多个单词的行，例如序列。这是做一个顺序模式挖掘分析。

我已经尝试过distinct()andduplicated()功能，但这只会删除整行。

r_seq_5 <- r_seq_5[!duplicated(r_seq_5),] # remove duplicates


   #       Su Score result ROI       next_roi  third_roi  four_roi   five_roi   
   #  1     1    90 high   Elsewhere Elsewhere Teacher    Teacher    Teacher   
   #  2     1    90 high   Elsewhere Teacher   Teacher    Teacher    Teacher   
   #  3     1    90 high   Teacher   Pen       Teacher    Elsewhere  Smartboard

这是桌子。如果Teacher在句子中出现了两三次也没关系，只要不是一个接一个就行。

期望的结果是：

# 1     1    90 high   Teacher   Pen       Teacher    Elsewhere  Smartboard
4

2 回答 2

0

您可以使用gather()为了重新组合您的变量，然后构建一个循环来识别与前一个相同的值。

最后，使用spread()以重建您的初始结构。

df <- data.frame(
  row = 1:4,
  Su = 1,
  Score = 90,
  result = 'high',
  ROI  = c('A', 'A', 'B', 'A'),
  ROI2 = c('A', 'B', 'C', 'B'),
  ROI3 = c('B', 'B', 'A', 'C')
) %>% 
  gather(-(row:result), key = roi, value = value) %>% 
  arrange(row) %>% 
  mutate(repeated = 0)

for(i in 2:nrow(df)){
  if(df$row[i] == df$row[i-1] & df$value[i] == df$value[i-1])
    df$repeated[i] = 1
}

df %>% 
  group_by(row) %>% 
  mutate(repeated = sum(repeated)) %>% 
  filter(repeated == 0) %>% 
  select(-repeated) %>% 
  spread(key = roi, value = value)

#     row    Su Score result ROI   ROI2  ROI3 
#   <int> <dbl> <dbl> <fct>  <chr> <chr> <chr>
# 1     3     1    90 high   B     C     A    
# 2     4     1    90 high   A     B     C
于 2019-05-13T15:19:21.317 回答
0

为此，我发现将因子转换为数字很方便。这是我的第一步，因为要比较列的数量，这条路径似乎不那么艰巨。

为此，我使用了 afor包qdap，因为在 macth 中我将值替换为NA.

library(dplyr)
library(qdap)
df <- data.frame(Su = rep(1,3),
                 Score = rep(90,3),
                 ROI = c("A", "A", "B"),
                 NETX_ROI = c("A", "B", "C"),
                 third_roi = rep("B", 3),
                 four_roi = c("B", "B", "A"),
                 five_roi = c("B", "B", "D"))
df

> df
  Su Score ROI NETX_ROI third_roi four_roi five_roi
1  1    90   A        A         B        B        B
2  1    90   A        B         B        B        B
3  1    90   B        C         B        A        D

df2 <- df
roi <- c("A", "B", "C", "D")
# A = Elsewhere
# B = Teacher
# C = Pen
# D = Smartboard

n <- seq(1, length.out = length(roi))
for (i in 1:length(n)) {
  df2[df2 == roi[i]] <- NA
  df2 <- qdap::NAer(df2, i)
}

> df2
  Su Score ROI NETX_ROI third_roi four_roi five_roi
1  1    90   1        1         2        2        2
2  1    90   1        2         2        2        2
3  1    90   2        3         2        1        4

df2 <- df2 %>% 
  dplyr::select(-c(Su, Score)) %>% 
  as.matrix()

nn <- ncol(df2)
x  <- matrix(nrow = nrow(df2), ncol = ncol(df2)-1)
for (i in 1:(nn-1)) {
  xx <- ifelse(df2[,i] == df2[,i+1], NA, 0)
  x[,i] <- as.matrix(xx)
}

> x
     [,1] [,2] [,3] [,4]
[1,]   NA    0   NA   NA
[2,]    0   NA   NA   NA
[3,]    0    0    0    0

最后，我只是删除了带有NA.

dfx <- x %>% 
  as.data.frame()

df_test <- df %>% 
  dplyr::bind_cols(dfx) %>% 
  na.omit() %>% 
  dplyr::select(1:ncol(df))
df_test

> df_test
  Su Score ROI NETX_ROI third_roi four_roi five_roi
3  1    90   B        C         B        A        D
于 2019-05-13T16:18:11.393 回答