Question

原始数据：

TRANS_DATE  ID          Value
10/2/2014   CPSG_CPM    2765.98
10/3/2014   CPSG_CPM    2840.76
10/6/2014   CPSG_CPM    3009.83
10/7/2014   CPSG_CPM    3025.05
10/8/2014   CPSG_CPM    2997.1
10/9/2014   CPSG_CPM    2946.08
10/10/2014  CPSG_CPM    2977.12
10/13/2014  CPSG_CPM    2797.95
10/14/2014  CPSG_CPM    2805.27
10/15/2014  CPSG_CPM    2768.37
10/16/2014  CPSG_CPM    2699.4
10/17/2014  CPSG_CPM    2841.46
10/20/2014  CPSG_CPM    2876.85

我想添加另一个列，它会为我提供上一个日期值的减去信息。

TRANS_DATE  ID          Value     Diff
10/2/2014   CPSG_CPM    2765.98   74.78
10/3/2014   CPSG_CPM    2840.76   169.07
10/6/2014   CPSG_CPM    3009.83   15.22
10/7/2014   CPSG_CPM    3025.05   -27.95
10/8/2014   CPSG_CPM    2997.1    -51.02
10/9/2014   CPSG_CPM    2946.08   31.04
10/10/2014  CPSG_CPM    2977.12   -179.17
10/13/2014  CPSG_CPM    2797.95   7.32
10/14/2014  CPSG_CPM    2805.27   -36.9
10/15/2014  CPSG_CPM    2768.37   -68.97
10/16/2014  CPSG_CPM    2699.4    142.06
10/17/2014  CPSG_CPM    2841.46   35.39
10/20/2014  CPSG_CPM    2876.85

Answer 1

不是最优雅的解决方案，但它确实有效。 dx cbind需要将{0}作为data.frame的最后一个元素追加为零{/ 1}}

d = as.Date(c("10/2/2014","10/3/2014","10/6/2014","10/7/2014"),format="%m/%d/%Y")
x = c(2765.98,2840.76,3009.83,3025.05)

dx = diff(x)
dx = append(dx,0)
print(dx)

df = data.frame(d,x,dx)
View(df)

    d            x       dx
1   2014-10-02  2765.98 74.78
2   2014-10-03  2840.76 169.07
3   2014-10-06  3009.83 15.22
4   2014-10-07  3025.05 0.00

如何减去R中前一天的值行？

1 个答案: