Question

如下面的df

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预期输出

我需要将df <- data.frame( name = rep(c("A", "B", "C"),2), type = c("10", "10", "10","20", "20", "20"), val = c(1,2,3,4,5,6) ) > df name type val 1 A 10 1 2 B 10 2 3 C 10 3 4 A 20 4 5 B 20 5 6 C 20 6 > val的所有记录中的name添加到C name的记录的val中，用于相应的A一个新的type AC。需要保留name name并且没有它的输出。

OUTPUT1

OUTPUT2

  name type val
1    A   10   1
2    B   10   2
3    C   10   3
4    AC  10   4
5    A   20   4
6    B   20   5
7    C   20   6
8    AC  20   10

更喜欢基于name type val 1 AC 10 4 2 B 10 2 4 AC 20 10 5 B 20 5 >的解决方案

Answer 1

这是一种方式，

shared_ptr

然后获得其他输出，

library(dplyr)

df %>% 
  mutate(new = as.integer(name %in% c('A', 'C'))) %>% 
  group_by(type, new) %>% 
  summarise(name = paste0(name, collapse = ''), val = sum(val)) %>% 
  ungroup() %>% 
  select(-new)

# A tibble: 4 × 3
#    type  name   val
#  <fctr> <chr> <dbl>
#1     10     B     2
#2     10    AC     4
#3     20     B     5
#4     20    AC    10

Answer 2

这是另一个（需要tidyr以及dplyr）

df1 <- df %>% group_by(type) %>% 
              summarise(AC=sum(val[name %in% c("A","C")]),B=val[name=="B"]) %>% 
              gather(key=name,value=val,-type) %>% 
              arrange(type)

Answer 3

以下是使用int input_number(const char* question) { int k = 0; char buffer[100]; printf("%s", question); fgets(buffer, sizeof buffer, stdin); sscanf(buffer, "%d", &k); return(k); }

的一个选项

data.table

或library(data.table) rbindlist(list(df, setDT(df)[, .(name = "AC", val = sum(val[as.character(name) %chin% c("A", "C")])) , .(type)][, names(df), with = FALSE]))[order(type, name)] # name type val #1: A 10 1 #2: B 10 2 #3: C 10 3 #4: AC 10 4 #5: A 20 4 #6: B 20 5 #7: C 20 6 #8: AC 20 10

dplyr

在条件上跨行添加列

3 个答案: