R 如果值大于3倍，如何将下一个重复值设置为空白
我有一个数据框，如下所示。在数据框中，值45重复/出现的次数大于3次，值67重复/出现的次数也相同，对于重复/冻结的次数大于3次的新值，现在需要将其设为空白/NAR 如果值大于3倍，如何将下一个重复值设置为空白,r,R,我有一个数据框，如下所示。在数据框中，值45重复/出现的次数大于3次，值67重复/出现的次数也相同，对于重复/冻结的次数大于3次的新值，现在需要将其设为空白/NA Name Value New_Value A 24 24 A 45 45 A 45 A 45 A 45 A 45 A 93 93 A
Name Value New_Value
A 24 24
A 45 45
A 45
A 45
A 45
A 45
A 93 93
A 19 19
A 10 10
B 29 29
B 67 67
B 67
B 67
B 67
C 201 201
C 993 993
C 396 396
library(data.table)
dt <- data.table(Names = LETTERS[1:5] %>% sample(100, replace = TRUE),
Value = sample(1:10, 100, replace = TRUE))
dt <- dt[order(Names, Value)]
# if you look for in-group duplicates
dt[, count := .N, by = .(Names, Value)][, New_Value := Value]
dt[ , dup_ingroup := duplicated(Value), by = Names]
dt[dup_ingroup & count > 3, New_Value := NA]
# if you look for all duplicates
dt[, count := .N, by = Value][, New_Value := Value]
dt[duplicated(Value) & count > 3, New_Value := NA]
library(data.table)
library(dplyr)
set.seed(20170515)
dt <- data.table(Names = LETTERS[1:5] %>% sample(100, replace = TRUE),
Value = sample(1:10, 100, replace = TRUE))
dt <- dt[order(Names, Value)]
dt_1 <- copy(dt)
dt_2 <- copy(dt)
dt_Jaap <- copy(dt)
# Method 1
dt_1[, count := .N, by = .(Names, Value)][, New_Value := Value]
dt_1[ , dup_ingroup := duplicated(Value), by = Names]
dt_1[dup_ingroup & count > 3, New_Value := NA]
dt_1[, .N, by = is.na(New_Value)]
## is.na N
## 1: FALSE 73
## 2: TRUE 27
# Method 2
dt_2[, count := .N, by = Value][, New_Value := Value]
dt_2[duplicated(Value) & count > 3, New_Value := NA]
dt_2[, .N, by = is.na(New_Value)]
## is.na N
## 1: FALSE 12
## 2: TRUE 88
# Method suggested by @Jaap
dt_Jaap[, New_Value := Value][duplicated(Value) & .N > 3, New_Value := NA_integer_, by = .(Names, Value)]
dt_Jaap[, .N, by = is.na(New_Value)]
## is.na N
## 1: FALSE 10
## 2: TRUE 90
df <- data.frame(Name = c("A","A","A","A","A","A","A","A","A","B","B","B","B","B","C","C","C","C","C"),
Value = c(24,45,45,45,45,45,93,19,10,29,67,67,67,67,201,993,396,396,396),
stringsAsFactors = F)
library(dplyr)
df %>%
group_by(Name, Value) %>%
mutate(New_Value = ifelse(n() > 3 & row_number() > 1, NA, Value))
df <- read.table(header = T, stringsAsFactors = F, text = "
Name Value
A 45
A 45
A 45
A 82
A 45
A 45
A 45
A 45
A 12
A 45
A 45
A 45
A 45
A 45
B 29
B 67
B 67
B 67
B 67
")
library(dplyr)
df %>%
group_by(Name) %>%
mutate(run_length = with(rle(Value), rep(lengths, lengths))) %>%
mutate(run_start = seq_along(Value) %in% cumsum(c(1, rle(Value)$lengths))) %>%
mutate(New_Value = ifelse(run_length < 4 | run_start, Value, NA)) %>%
ungroup() %>% select(-run_length, -run_start)
DT
# Name Value
# 1: A 24
# 2: A 24
# 3: A 45
# 4: A 45
# 5: A 45
# 6: A 45
# 7: A 45
# 8: A 93
# 9: A 19
#10: A 19
#11: A 10
#12: B 29
#13: B 67
#14: B 67
#15: B 67
#16: B 67
#17: C 201
#18: C 993
#19: C 396
#20: A 19
#21: A 19
#22: C 19
#23: B 29
#24: B 67
#25: B 67
#26: B 67
#27: B 67
#28: C 67
#29: C 67
#30: C 67
#31: C 67
# Name Value
library(data.table)
setDT(DT)[, New := Value[.N < 3], by=rleid(Value)][rowid(rleid(Value)) == 1L, New := Value]
DT
# Name Value New
# 1: A 24 24
# 2: A 24 24
# 3: A 45 45
# 4: A 45 NA
# 5: A 45 NA
# 6: A 45 NA
# 7: A 45 NA
# 8: A 93 93
# 9: A 19 19
#10: A 19 19
#11: A 10 10
#12: B 29 29
#13: B 67 67
#14: B 67 NA
#15: B 67 NA
#16: B 67 NA
#17: C 201 201
#18: C 993 993
#19: C 396 396
#20: A 19 19
#21: A 19 NA
#22: C 19 NA
#23: B 29 29
#24: B 67 67
#25: B 67 NA
#26: B 67 NA
#27: B 67 NA
#28: C 67 NA
#29: C 67 NA
#30: C 67 NA
#31: C 67 NA
# Name Value New
setDT(DT)[, New := Value[.N < 3], by = rleid(Name, Value)
][is.na(New) & rowid(rleid(Name, Value)) == 1L, New := Value][]
# Name Value New
# 1: A 24 24
# 2: A 24 24
# 3: A 45 45
# 4: A 45 NA
# 5: A 45 NA
# ...
#18: C 993 993
#19: C 396 396
#20: A 19 19
#21: A 19 19
#22: C 19 19
#23: B 29 29
#24: B 67 67
#25: B 67 NA
#26: B 67 NA
#27: B 67 NA
#28: C 67 67
#29: C 67 NA
#30: C 67 NA
#31: C 67 NA
# Name Value New
请注意，第1行和第8行已被重复。OP的数据集覆盖了两次重复的情况，并且在末尾添加了一个fews行。这里已经回答了这个问题：可能重复@Larusson，这不是重复的问题。海报需要替换显示超过3倍的数据值，所有答案都涉及重复的普通答案或数据。表中未说明顺序重复的问题。从请求中也不清楚提问者是否理解这个问题。如果一个值恰好重复两次，你会怎么想？不幸的是，您的示例数据不包括这种情况。您可以将其缩短。对于组内重复项：dt[，New_Value:=Value][duplicatedValue&.N>3，New_Value:=NA_integer，by=.Name，Value]。如您所见，您不必创建count变量，您可以直接在i.@Jaap中使用.N，这似乎不起作用，因为duplicatedValue&.N>3部分没有按变量分组。dt[，.N，by=is.naNew_值]的结果对于我的和你的不一样。它给了我正确的输出。您是否已更新到data.table v1.10.4的最新版本？很抱歉误解，您是正确的。我原以为.N现在可以用在I中，但我在新闻文件中找不到它；显然情况并非如此：-/修复了代码中的一个输入错误。。。使用数据而不是df作为数据帧名称，但从概念上讲，这正是相同的请注意，如果相同的值重复两次，您的解决方案还将为第二个值添加NA。OP要求仅当值重复三次或更多次时才出现空白。不幸的是，OP的样本数据集不包括两次重复的情况。我在第一次阅读时不知何故错过了>3次。我更新了代码，假设数据帧中的行顺序不重要，我认为：=v[logi]是一件奇怪的事情。当logi为false时，RHS的长度为零，我希望这会破坏一切。。。如果没有，那么在我看来，这依赖于一个奇怪的实现怪癖。不管怎样，我想我希望看到x:=如果logi v else v[NA_integer_u]@Frank自1.9.8版以来，这不再奇怪：即使没有匹配项或其RHS为长度0，新列也保证为：=参见中的第二项。感谢指针。嗯，古怪是主观的：出于同样的原因，我喜欢logi.SD超过.SD[logi]。一般来说，当出现错误时，这两个奇怪/怪异语法方面的问题都会消失
setDT(DT)[, New := Value[.N < 3], by = rleid(Name, Value)
][is.na(New) & rowid(rleid(Name, Value)) == 1L, New := Value][]
# Name Value New
# 1: A 24 24
# 2: A 24 24
# 3: A 45 45
# 4: A 45 NA
# 5: A 45 NA
# ...
#18: C 993 993
#19: C 396 396
#20: A 19 19
#21: A 19 19
#22: C 19 19
#23: B 29 29
#24: B 67 67
#25: B 67 NA
#26: B 67 NA
#27: B 67 NA
#28: C 67 67
#29: C 67 NA
#30: C 67 NA
#31: C 67 NA
# Name Value New
DT <- structure(list(Name = c("A", "A", "A", "A", "A", "A", "A", "A",
"A", "A", "A", "B", "B", "B", "B", "B", "C", "C", "C", "A", "A",
"C", "B", "B", "B", "B", "B", "C", "C", "C", "C"), Value = c(24L,
24L, 45L, 45L, 45L, 45L, 45L, 93L, 19L, 19L, 10L, 29L, 67L, 67L,
67L, 67L, 201L, 993L, 396L, 19L, 19L, 19L, 29L, 67L, 67L, 67L,
67L, 67L, 67L, 67L, 67L)), .Names = c("Name", "Value"), row.names = c(NA,
-31L), class = "data.frame")