基于特定条件过滤和添加的有效方法(本例中为 3 个条件) [英] Efficient method to filter and add based on certain conditions (3 conditions in this case)

问题描述

我有一个看起来像这样的数据框

I have a data frame which looks like this

     a    b    c   d
     1    1    1   0
     1    1    1   200
     1    1    1   300
     1    1    2   0
     1    1    2   600
     1    2    3   0
     1    2    3   100
     1    2    3   200
     1    3    1   0

我有一个看起来像这样的数据框

I have a data frame which looks like this

     a    b    c   d
     1    1    1   250
     1    1    2   600
     1    2    3   150
     1    3    1   0

我现在正在做{

  n=nrow(subset(Wallmart, a==i &    b==j & c==k  ))
  sum=subset(Wallmart, a==i &    b==j & c==k  )
  #sum
  sum1=append(sum1,sum(sum$d)/(n-1))

}

我想添加d"列并通过计算行数而不计算 0 来取平均值.例如第一行是 (200+300)/2 = 250.目前我正在构建一个存储d"列的列表，但理想情况下我希望它采用上述格式.例如第一行看起来像

I would like to add the 'd' coloumn and take the average by counting the number of rows without counting 0. For example the first row is (200+300)/2 = 250. Currently I am building a list that stores the 'd' coloumn but ideally I want it in the format above. For example first row would look like

     a    b    c   d
     1    1    1   250

这是一种非常低效的方式来完成这项工作.代码需要很长时间才能循环运行.因此，感谢任何帮助，使其运行得更快.原始数据框有大约一百万行.

This is a very inefficient way to do this work. The code takes a long time to run in a loop. so any help is appreciated that makes it run faster. The original data frame has about a million rows.

推荐答案

你可以试试aggregate:

aggregate(d ~ a + b + c, data = df, sum)
#   a b c   d
# 1 1 1 1 500
# 2 1 3 1   0
# 3 1 1 2 600
# 4 1 2 3 300

正如@Roland 所说，对于更大的数据集，您可以尝试 data.table 或 dplyr，例如:

As noted by @Roland, for bigger data sets, you may try data.table or dplyr instead, e.g.:

library(dplyr)
df %>%
  group_by(a, b, c) %>%
  summarise(
    sum_d = sum(d))

# Source: local data frame [4 x 4]
# Groups: a, b
# 
#   a b c sum_d
# 1 1 1 1   500
# 2 1 1 2   600
# 3 1 2 3   300
# 4 1 3 1     0

编辑更新问题.如果你想计算分组平均值，不包括零行，你可以试试这个:

Edit following updated question. If you want to calculate group-wise mean, excluding rows that are zero, you may try this:

aggregate(d ~ a + b + c, data = df, function(x) mean(x[x > 0]))
#   a b c   d
# 1 1 1 1 250
# 2 1 3 1 NaN
# 3 1 1 2 600
# 4 1 2 3 150

df %>%
  filter(d != 0) %>%
  group_by(a, b, c) %>%
  summarise(
    mean_d = mean(d))

#   a b c mean_d
# 1 1 1 1    250
# 2 1 1 2    600
# 3 1 2 3    150

但是，因为您似乎希望将零视为缺失值而不是数字零，所以我认为在准备数据集时将它们转换为 NA 会更好，在计算之前.

However, because it seems that you wish to treat your zeros as missing values rather than numeric zeros, I think it would be better to convert them to NA when preparing your data set, before the calculations.

df$d[df$d == 0] <- NA
df %>%
  group_by(a, b, c) %>%
  summarise(
    mean_d = mean(d, na.rm = TRUE))

#   a b c mean_d
# 1 1 1 1    250
# 2 1 1 2    600
# 3 1 2 3    150
# 4 1 3 1    NaN

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