Question

我有一个包含2列和3659行df

的数据框

我试图通过平均此数据框中的每10行或13行来减少数据集，所以我尝试了以下内容：

# number of rows per group
n=13
# number of groups
n_grp=nrow(df)/n
round(n_grp,0)
# row indices (one vector per group)
idx_grp <- split(seq(df), rep(seq(n_grp), each = n))

# calculate the col means for all groups
res <- lapply(idx_grp, function(i) {
  # subset of the data frame
  tmp <- dat[i]
  # calculate row means
  colMeans(tmp, na.rm = TRUE)
})
# transform list into a data frame
dat2 <- as.data.frame(res)

但是，我不能将行数除以10或13，因为数据长度不是拆分变量的倍数。所以我不确定应该做什么（我只想要计算最后一组的平均值 - 即使少于10个元素）

我也试过这个，但结果是一样的：

df1=split(df, sample(rep(1:301, 10)))

Answer 1

以下是使用aggregate()和rep()的解决方案。

df <- data.frame(a=1:12, b=13:24 );
df;
##     a  b
## 1   1 13
## 2   2 14
## 3   3 15
## 4   4 16
## 5   5 17
## 6   6 18
## 7   7 19
## 8   8 20
## 9   9 21
## 10 10 22
## 11 11 23
## 12 12 24
n <- 5;
aggregate(df,list(rep(1:(nrow(df)%/%n+1),each=n,len=nrow(df))),mean)[-1];
##      a    b
## 1  3.0 15.0
## 2  8.0 20.0
## 3 11.5 23.5

此解决方案的一个重要部分是nrow(df) n len处理length.out的不可分性问题，指定rep()参数（实际上参数名称为ResultSet resSet; resSet = statement.executeQuery("SELECT COUNT(*) FROM table"); resSet.next() long rowCount = resSet.getLong(1); resSet = statement.executeQuery("SELECT * FROM table"); // read data of known row count...} ）connection.setAutoCommit(false); connection.setTransactionIsolation(Connection.TRANSACTION_SERIALIZABLE) // call 2 SQL queries above connection.commit();，它自动将组矢量限制为适当的长度。

Answer 2

如果df是data.table，您可以使用%/%分组，如

library(data.table)
setDT(df)
n <- 13 # every 13 rows

df[, mean(z), by= (seq(nrow(df)) - 1) %/% n]

如果您想要每个第n行，请使用%%代替%/%

df[, mean(z), by= (seq(nrow(df)) - 1) %% n]

Answer 3

这应该有效。使用n = 13将13行聚集在一起。如果您有27行，则会获得大小为13,13,1的组。

n.colmeans = function(df, n = 10){
    aggregate(x = df,
              by = list(gl(ceiling(nrow(df)/n), n)[1:nrow(df)]),
              FUN = mean)
}

n.colmeans(state.x77, 10)

  Group.1 Population Income Illiteracy Life Exp Murder HS Grad Frost     Area
1       1     4892.8 4690.8       1.44   70.508   9.53   53.63  75.1 116163.6
2       2     3570.5 4419.4       1.12   71.110   7.07   53.35  99.8  44406.6
3       3     3697.9 4505.5       0.93   70.855   6.64   55.25 131.7  60873.0
4       4     5663.9 4331.2       1.33   70.752   7.12   49.59 103.6  56949.5
5       5     3407.0 4232.1       1.03   71.168   6.53   53.72 112.1  75286.7

Answer 4

dplyr 路

n1 <- 10
iris %>% group_by(mean = (row_number() -1) %/% n1) %>%
  mutate(mean = mean(Sepal.Length))

# A tibble: 150 x 6
# Groups:   mean [15]
   Sepal.Length Sepal.Width Petal.Length Petal.Width Species  mean
          <dbl>       <dbl>        <dbl>       <dbl> <fct>   <dbl>
 1          5.1         3.5          1.4         0.2 setosa   4.86
 2          4.9         3            1.4         0.2 setosa   4.86
 3          4.7         3.2          1.3         0.2 setosa   4.86
 4          4.6         3.1          1.5         0.2 setosa   4.86
 5          5           3.6          1.4         0.2 setosa   4.86
 6          5.4         3.9          1.7         0.4 setosa   4.86
 7          4.6         3.4          1.4         0.3 setosa   4.86
 8          5           3.4          1.5         0.2 setosa   4.86
 9          4.4         2.9          1.4         0.2 setosa   4.86
10          4.9         3.1          1.5         0.1 setosa   4.86
# ... with 140 more rows

或者如果 n1 不是除数或 nrow(df) 那么也

n1 <- 7

# A tibble: 150 x 6
# Groups:   mean [21]
   Sepal.Length Sepal.Width Petal.Length Petal.Width Species  mean
          <dbl>       <dbl>        <dbl>       <dbl> <fct>   <dbl>
 1          5.1         3.5          1.4         0.2 setosa    4.9
 2          4.9         3            1.4         0.2 setosa    4.9
 3          4.7         3.2          1.3         0.2 setosa    4.9
 4          4.6         3.1          1.5         0.2 setosa    4.9
 5          5           3.6          1.4         0.2 setosa    4.9
 6          5.4         3.9          1.7         0.4 setosa    4.9
 7          4.6         3.4          1.4         0.3 setosa    4.9
 8          5           3.4          1.5         0.2 setosa    4.8
 9          4.4         2.9          1.4         0.2 setosa    4.8
10          4.9         3.1          1.5         0.1 setosa    4.8
# ... with 140 more rows

您还可以跨多个列进行变异

mydf <- iris[-5]

mydf %>% group_by(n = (row_number() -1) %/% n1) %>%
  mutate(across(everything(), ~ mean(.), .names = "{.col}_mean"))

# A tibble: 150 x 9
# Groups:   n [22]
   Sepal.Length Sepal.Width Petal.Length Petal.Width     n Sepal.Length_me~ Sepal.Width_mean Petal.Length_me~
          <dbl>       <dbl>        <dbl>       <dbl> <dbl>            <dbl>            <dbl>            <dbl>
 1          5.1         3.5          1.4         0.2     0              4.9             3.39             1.44
 2          4.9         3            1.4         0.2     0              4.9             3.39             1.44
 3          4.7         3.2          1.3         0.2     0              4.9             3.39             1.44
 4          4.6         3.1          1.5         0.2     0              4.9             3.39             1.44
 5          5           3.6          1.4         0.2     0              4.9             3.39             1.44
 6          5.4         3.9          1.7         0.4     0              4.9             3.39             1.44
 7          4.6         3.4          1.4         0.3     0              4.9             3.39             1.44
 8          5           3.4          1.5         0.2     1              4.8             3.21             1.43
 9          4.4         2.9          1.4         0.2     1              4.8             3.21             1.43
10          4.9         3.1          1.5         0.1     1              4.8             3.21             1.43
# ... with 140 more rows, and 1 more variable: Petal.Width_mean <dbl>

计算数据框中每13行的平均值

4 个答案: