我这里有数据表：
row V1 velocity
1 2009-04-06 95.9230769230769
2 2009-04-11 95.0985074626866
3 2009-04-17 95.8064935064935
4 2009-04-22 94.6357142857143
5 2009-04-27 95.3626865671642
6 2009-05-03 95.9101265822785
7 2009-05-08 95.826582278481
8 2009-05-14 94.5126582278481
9 2009-05-20 95.8371428571429
10 2009-05-25 94.6981481481481
11 2009-05-30 96.397619047619
12 2009-06-05 94.8132530120482
13 2009-06-10 96.4558139534884
14 2009-06-16 94.9627906976744
15 2009-06-21 95.2666666666667
16 2009-06-26 95.2919540229885
17 2009-07-01 95.4333333333333
18 2009-07-07 95.3375
19 2009-07-12 95.0534246575343
20 2009-07-18 96.0277777777778
21 2009-07-24 95.6885057471264
22 2009-07-29 93.9375
23 2009-08-03 95.2776315789474
24 2009-08-08 94.9089285714286
25 2009-08-13 96.8906976744186
26 2009-08-19 95.4487804878049
27 2009-08-24 97.2444444444444
28 2009-08-30 95.1174418604651
我想写一个r代码来找到每月的速度平均值。 （有五月，六月，七月和八月。
我该怎么办？
答案 0 :(得分：3)
或者jusr：
tapply(df$velocity, months(as.Date(df$V1)), mean)
April August Juli Juni Mai
95.36530 95.81465 95.24634 95.35810 95.53038
答案 1 :(得分：2)
以下是我将如何做到这一点
使用lubridate创建一个月份变量，以便在dplyr中分组，然后获取资格。
library(lubridate)
library(dplyr)
df %>% group_by(month = month(df$V1)) %>% summarize(mean = mean(velocity))
month mean
1 4 95.36530
2 5 95.53038
3 6 95.35810
4 7 95.24634
5 8 95.81465
如果你添加label=T，你会得到这个：
df %>% group_by(month = month(df$V1,label=T)) %>% summarize(mean = mean(velocity))
month mean
1 Apr 95.36530
2 May 95.53038
3 Jun 95.35810
4 Jul 95.24634
5 Aug 95.81465