按组计算连续行中的值之间的差异 [英] Calculate difference between values in consecutive rows by group

问题描述

这是我的df (data.frame):

This is a my df (data.frame):

group value
1     10
1     20
1     25
2     5
2     10
2     15 

我需要按组计算连续行中的值之间的差异.

I need to calculate difference between values in consecutive rows by group.

所以，我需要一个结果.

So, I need a that result.

group value diff
1     10    NA # because there is a no previous value
1     20    10 # value[2] - value[1]
1     25    5  # value[3] value[2]
2     5     NA # because group is changed
2     10    5  # value[5] - value[4]
2     15    5  # value[6] - value[5]

虽然我可以通过使用ddply来处理这个问题，但是太费时间了.这是因为我的 df 中有很多组.(我的 df 中有超过 1,000,000 个组)

Although, I can handle this problem by using ddply, but it takes too much time. This is because I have a lot of groups in my df. (over 1,000,000 groups in my df)

有没有其他有效的方法来处理这个问题?

Are there any other effective approaches to handle this problem?

推荐答案

包 data.table 可以使用 shift 函数相当快地做到这一点.

The package data.table can do this fairly quickly, using the shift function.

require(data.table)
df <- data.table(group = rep(c(1, 2), each = 3), value = c(10,20,25,5,10,15))
#setDT(df) #if df is already a data frame

df[ , diff := value - shift(value), by = group]    
#   group value diff
#1:     1    10   NA
#2:     1    20   10
#3:     1    25    5
#4:     2     5   NA
#5:     2    10    5
#6:     2    15    5
setDF(df) #if you want to convert back to old data.frame syntax

<小时>

或者使用dplyr

df %>%
    group_by(group) %>%
    mutate(Diff = value - lag(value))
#   group value  Diff
#   <int> <int> <int>
# 1     1    10    NA
# 2     1    20    10
# 3     1    25     5
# 4     2     5    NA
# 5     2    10     5
# 6     2    15     5

<小时>

对于 pre-data.table::shift 和 pre-dplyr::lag 的替代方案，请参阅编辑.

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For alternatives pre-data.table::shift and pre-dplyr::lag, see edits.

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