Question

我有一个数据框如下。在数据框中，对于“A”重复/出现的值“45”大于3次，对于“B”对于“67”也是相同的，现在需要将它们作为“空白/ NA”用于重复/冻结的那些大于3次（“New_value”）

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

Answer 1

FWIW，这是使用data.table而非rleid()的另一个duplicated()解决方案。

请注意，如果值显示大于3次，则OP请求将下一个重复值设为空白。这意味着对于Value重复两次没有空白应该出现在结果中。我修改了我的样本数据集，以包含两次重复相同值的情况。

编辑：OP没有明确表示他是否正在计算给定序列中相同Value的重复次数，而不管Name是否为Name计算每个Value组序列中的重复次数。另见this comment。

此外，OP没有指定他预期的结果，如果有一系列重复Name但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发生了变化。

因此，我修改了我的示例数据集以包含其他用例：

NA

与其他答案一样，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为空白。

Value

第一个表达式为所有RLE组复制NA，重复一次或两次。重复次数更多的所有RLE组获得Value。第二个表达式仅为每个RLE组中的第一行复制Name。

请注意，重复值的每个序列都是单独处理的，无论A是什么，但第22行C到B和C到setDT(DT)[, New := Value[.N < 3], by=rleid(Value) ][is.na(New) & rowid(rleid(Value)) == 1L, New := Value]的变化第27行被忽略了。

这可以进一步改进，只有在尚未复制时才能复制：

Name

如果Value的变化预期为＆＃34;重新启动＆＃34; 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也可以使用这个变体（Jaap的信用）：

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")

注意第21,22和27行的差异。

数据

;WITH tbl_to_synch as ( -- Prepare table to update, Select *,chk = CHECKSUM(*) from [dbo].[tableA] ) MERGE tbl_to_synch as [Target] USING (Select *,chk = CHECKSUM(*) from [dbo].[tableB]) as [source] ON [Target].key = [source].key WHEN MATCHED AND [Target].chk <> [source].chk THEN -- UPDATE ONLY row that is changed UPDATE SET column01 = [source].[column01] ,column02 = [source].[column01] -- .... ,column59 = [source].[column59] ,column60 = [source].[column59] WHEN NOT MATCHED BY TARGET THEN insert (column01, column02, ...,column59,column60) values (column01, column02, ...,column59,column60) WHEN NOT MATCHED BY SOURCE THEN DELETE -- Show what is changed OUTPUT $action, ISNULL(INSERTED.key,DELETED.key);

请注意，第1行和第8行已经重复了w.r.t. OP的数据集覆盖了两次重复的情况，最后添加了几行。

Answer 2

这是data.table方法。我为

展示了两种解决方案

您在Names组
您在所有数据中查找重复项

以下是代码：

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]

注意

以下评论。

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

dt_Jaap只保留每个值Value的第一个元素的值。

Answer 3

和dplyr / tidyverse方式，假设数据框的顺序无关紧要......

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))

<强>更新

一种更强大的方法，可以处理多组相同的值...

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)

如果值显示大于3次，如何将下一个重复值作为空白

3 个答案:

数据