计算数据框中每 13 行的平均值 [英] Calculate the mean of every 13 rows in data frame
问题描述
我有一个 2 列 3659 行的数据框 df
I have a data frame with 2 columns and 3659 row df
我试图通过在此数据框中每 10 或 13 行求平均值来减少数据集,因此我尝试了以下操作:
I am trying to reduce the data set by averaging every 10 or 13 rows in this data frame, so I tried the following :
# 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.所以我不确定应该怎么做(我只想计算最后一组的平均值 - 即使少于 10 个元素)
However, I can't divide my number of rows by 10 or 13 because data length is not a multiple of split variable. So I am not sure what should do then (I just want may be to calculate the mean of the last group -even with less than 10 elements)
我也试过这个,但结果是一样的:
I also tried this one, but the results are the same:
df1=split(df, sample(rep(1:301, 10)))
推荐答案
这是一个使用 aggregate()
和 rep()
的解决方案.
Here's a solution using aggregate()
and rep()
.
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
该解决方案处理 nrow(df)
不能被 n
整除的问题的重要部分是指定 len
参数(其实参数全名是length.out
)rep()
,它会自动将组向量限制为合适的长度.
The important part of this solution that handles the issue of non-divisibility of nrow(df)
by n
is specifying the len
parameter (actually the full parameter name is length.out
) of rep()
, which automatically caps the group vector to the appropriate length.
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