我有一个数据框,有两个数字变量fatcontent和saltcontent加上两个因子变量cond和spice描述不同的处理.在该数据框中,对数值变量的每次测量取两次.
a <- data.frame(cond = rep(c("uncooked", "fried", "steamed", "baked", "grilled"),
each = 2, times = 3),
spice = rep(c("none", "chilli", "basil"), each = 10),
fatcontent = c(4, 5, 6828, 7530, 6910, 7132, 5885, 613, 2845, 2867,
25, 18, 2385, 33227, 4233, 4023, 953, 1025, 4465, 5016,
5, 5, 10235, 12545, 5511, 5111, 596, 585, 4012, 3633),
saltcontent = c(2, 5, 4733, 5500, 5724, 15885, 14885, 217, 193, 148,
6, 4, 26738, 24738, 22738, 23738, 267, 256, 1121, 1558,
1, 1, 21738, 20738, 26738, 27738, 195, 202, 129, 131)
)
现在,我希望通过未烹饪条件的平均值来表示每个香料组的数值变量(在这种情况下意味着除).
例如$ a spice =="none"
cond spice fatcontent saltcontent
1 uncooked none 4 2
2 uncooked none 5 5
3 fried none 6828 4733
4 fried none 7530 5500
5 steamed none 6910 5724
6 steamed none 7132 15885
7 baked none 5885 14885
8 baked none 613 217
9 grilled none 2845 193
10 grilled none 2867 148
正常化后:
cond spice fatcontent saltcontent
1 uncooked none 0.8888889 0.5714286
2 uncooked none 1.1111111 1.4285714
3 fried none 1517.3333333 1352.2857143
4 fried none 1673.3333333 1571.4285714
5 steamed none 1535.5555556 1635.4285714
6 steamed none 1584.8888889 4538.5714286
7 baked none 1307.7777778 4252.8571429
8 baked none 136.2222222 62.0000000
9 grilled none 632.2222222 55.1428571
10 grilled none 637.1111111 42.2857143
我的问题是如何为数据框中的所有组和变量执行此操作？我假设我可以使用dplyr包,但我不确定什么是最好的方法.我感谢任何帮助!
规范化数据的一种简洁方法是在均值计算中包含“未煮过的”条件，这样您就不需要过滤、汇总、连接和重新计算。这样做mutate_each意味着你只需要输入一次。
group_by(a, spice) %>%
mutate_each(funs(./mean(.[cond == "uncooked"])), -cond)
#Source: local data frame [30 x 4]
#Groups: spice
#
# cond spice fatcontent saltcontent
#1 uncooked none 0.8888889 5.714286e-01
#2 uncooked none 1.1111111 1.428571e+00
#3 fried none 1517.3333333 1.352286e+03
#4 fried none 1673.3333333 1.571429e+03
#5 steamed none 1535.5555556 1.635429e+03
#6 steamed none 1584.8888889 4.538571e+03
#7 baked none 1307.7777778 4.252857e+03
#8 baked none 136.2222222 6.200000e+01
#9 grilled none 632.2222222 5.514286e+01
#10 grilled none 637.1111111 4.228571e+01
# ... etc
我想这就是你所追求的。您想要使用未煮熟的数据点找到每种香料条件的平均值。这是我第一步所做的事情。然后，我想将fatmean和saltmean添加ana到您的数据框中，a. 如果您的数据确实很大，这可能不是一种有效利用内存的方法。但是，我曾经left_join合并ana和a。然后，我mutate对每种香料条件进行了划分。最后，我删除了两列来使用 整理结果select。
### Find mean for each spice condition using uncooked data points
ana <- group_by(filter(a, cond == "uncooked"), spice) %>%
summarise(fatmean = mean(fatcontent), saltmean = mean(saltcontent))
# spice fatmean saltmean
#1 basil 5.0 1.0
#2 chilli 21.5 5.0
#3 none 4.5 3.5
left_join(a, ana, by = "spice") %>%
group_by(spice) %>%
mutate(fatcontent = fatcontent / fatmean,
saltcontent = saltcontent / saltmean) %>%
select(-c(fatmean, saltmean))
# A part of the results
# cond spice fatcontent saltcontent
#1 uncooked none 0.8888889 0.5714286
#2 uncooked none 1.1111111 1.4285714
#3 fried none 1517.3333333 1352.2857143
#4 fried none 1673.3333333 1571.4285714
#5 steamed none 1535.5555556 1635.4285714
#6 steamed none 1584.8888889 4538.5714286
#7 baked none 1307.7777778 4252.8571429
#8 baked none 136.2222222 62.0000000
#9 grilled none 632.2222222 55.1428571
#10 grilled none 637.1111111 42.2857143
如果你在一个管道中完成所有的事情，它会是这样的：
group_by(filter(a, cond == "uncooked"), spice) %>%
summarise(fatmean = mean(fatcontent), saltmean = mean(saltcontent)) %>%
left_join(a, ., by = "spice") %>% #right_join is possible with the dev dplyr
group_by(spice) %>%
mutate(fatcontent = fatcontent / fatmean,
saltcontent = saltcontent / saltmean) %>%
select(-c(fatmean, saltmean))