在R中减去列

我有以下数据帧：

Name,2014,2015,2016,2017,2018
Brad,100,200,342,532,65
Tom,54,2,523,121,200
Uma,200,221,225,229,250

我想做的是每年从上一列中减去：

Name,2014,2015,2016,2017,2018
Brad,0,100,142,190,-467

我想自动完成，而不是像这样：

(df$"2003" - df$"2002")

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怎么可能呢？对不起，我是R的新手。谢谢

您可能想要

# this is your data
df <- data.frame("2014"=c(100,54,200),
             "2015"=c(200,2,221),
             "2016"=c(342,523,225),
             "2017"=c(532,121,229),
             "2018"=c(65,200,250),
             row.names=c("Brad","Tom","Uma"))

df     
#      X2014 X2015 X2016 X2017 X2018
# Brad   100   200   342   532    65
# Tom     54     2   523   121   200
# Uma    200   221   225   229   250

df[，-1]

给出data.frame中除第一列之外的每一列，

df[，-5]

给出除第五列之外的每一列。。。从这里开始，就是简单的减法

df <- structure(list(X2014 = c(100, 54, 200), X2015 = c(200, 2, 221
), X2016 = c(342, 523, 225), X2017 = c(532, 121, 229), X2018 = c(65, 
200, 250)), .Names = c("X2014", "X2015", "X2016", "X2017", "X2018"
), row.names = c("Brad", "Tom", "Uma"), class = "data.frame")

解决方案2 使用

dplyr

library(dplyr)
df <- data.frame(t(df)) %>% mutate_all(function(x) x - lag(x,default =x[1]))
df <- data.frame(t(df))

库（dplyr）
df%mutate_all（函数（x）x-lag（x，默认值=x[1]））
df
data.frame(t(apply(df, 1, diff)))

library(dplyr)
df <- data.frame(t(df)) %>% mutate_all(function(x) x - lag(x,default =x[1]))
df <- data.frame(t(df))