对多个“锚点”的R-重复函数
对多个“锚点”的R-重复函数
提问于 2016-07-20 04:51:10
很抱歉，如果这是一个措辞不佳的问题，我对用R组织数据的最佳方式感到困惑。以下是数据框架：
Experiment Treatment Inhibitor pAKT Loading Control
1 None DMSO 7075.426 16051.376
1 Carbachol DMSO 18368.962 14826.962
1 Carbachol 10-8 M BYL179 14949.184 10911.477
1 Carbachol 10-7 M BYL179 11972.477 10993.305
1 Carbachol 10-6 M BYL179 14874.841 12610.841
1 Carbachol 10-8 M Cal101 11433.598 12305.891
1 Carbachol 10-7 M Cal101 10297.719 13175.012
1 Carbachol 10-6 M Cal101 2835.891 11609.598
1 Carbachol 10-8 M BYL179 + Cal101 8749.134 12024.012
1 Carbachol 10-7 M BYL179 + Cal101 4490.841 13088.598
1 Carbachol 10-6 M BYL179 + Cal101 598.012 10929.134
2 None DMSO 11904.225 30755.104
2 Carbachol DMSO 27637.518 27536.912
2 Carbachol 10-8 M BYL179 13970.619 24052.326
2 Carbachol 10-7 M BYL179 7567.326 21889.154
2 Carbachol 10-6 M BYL179 8421.326 13446.033
2 Carbachol 10-8 M Cal101 4754.497 7333.912
2 Carbachol 10-7 M Cal101 3614.497 3566.376
2 Carbachol 10-6 M Cal101 1426.598 1331.326
2 Carbachol 10-8 M BYL179 + Cal101 7401.175 6010.447
2 Carbachol 10-7 M BYL179 + Cal101 3159.740 5280.154
2 Carbachol 10-6 M BYL179 + Cal101 2114.376 9793.347
3 None DMSO 4528.690 20969.347
3 Carbachol DMSO 27218.104 23433.912
3 Carbachol 10-8 M BYL179 12507.619 21712.033
3 Carbachol 10-7 M BYL179 4673.790 13934.154
3 Carbachol 10-6 M BYL179 6699.033 8018.497
3 Carbachol 10-8 M Cal101 10108.861 17364.861
3 Carbachol 10-7 M Cal101 10728.326 25307.447
3 Carbachol 10-6 M Cal101 7418.083 20826.012
3 Carbachol 10-8 M BYL179 + Cal101 17169.104 23850.255
3 Carbachol 10-7 M BYL179 + Cal101 14890.811 36585.368
3 Carbachol 10-6 M BYL179 + Cal101 5889.497 19609.033
4 None DMSO 5029.841 14217.083
4 EGF DMSO 17963.719 16883.255
4 EGF 10-8 M BYL179 18191.719 15706.719
4 EGF 10-7 M BYL179 17430.719 16327.891
4 EGF 10-6 M BYL179 15350.305 14828.134
4 EGF 10-8 M Cal101 15847.891 14122.598
4 EGF 10-7 M Cal101 17678.426 16659.719
4 EGF 10-6 M Cal101 14236.477 15380.012
4 EGF 10-8 M BYL179 + Cal101 14281.891 15367.012
4 EGF 10-7 M BYL179 + Cal101 13918.012 13796.062
4 EGF 10-6 M BYL179 + Cal101 4702.062 12578.012
5 None DMSO 14192.004 40446.196
5 EGF DMSO 43425.246 46626.125
5 EGF 10-8 M BYL179 43465.004 45206.832
5 EGF 10-7 M BYL179 34098.882 36853.175
5 EGF 10-6 M BYL179 33415.882 37637.539
5 EGF 10-8 M Cal101 32413.175 31047.054
5 EGF 10-7 M Cal101 32359.882 35967.004
5 EGF 10-6 M Cal101 27089.761 33929.125
5 EGF 10-8 M BYL179 + Cal101 30945.882 29804.882
5 EGF 10-7 M BYL179 + Cal101 32511.711 37228.832
5 EGF 10-6 M BYL179 + Cal101 10176.832 36218.024
6 None DMSO 17417.004 47193.439
6 EGF DMSO 49204.075 55932.489
6 EGF 10-8 M BYL179 45682.832 52486.125
6 EGF 10-7 M BYL179 37131.711 41015.004
6 EGF 10-6 M BYL179 32859.761 33375.933
6 EGF 10-8 M Cal101 39127.711 33037.004
6 EGF 10-7 M Cal101 35535.589 33514.539
6 EGF 10-6 M Cal101 25455.175 28547.882
6 EGF 10-8 M BYL179 + Cal101 21374.589 24441.296
6 EGF 10-7 M BYL179 + Cal101 19590.125 22268.660
6 EGF 10-6 M BYL179 + Cal101 5347.782 21943.317我想创建一个标题为“校正因子”的新列，通过将特定条目的that值除以该实验的处理为"None“和抑制剂为"DMSO”的条目而生成。因此，例如，实验1的第一个条目是16051.376/16051.376 = 1，第二个条目是14826.962/16051.376 = 0.9237，第三个条目是10911.477/16051.376 = 0.6798，依此类推。对于实验2，我将除以实验2无+ DMSO条目，30755.104。如何对每个实验重复此操作？
回答 1
Stack Overflow用户
发布于 2016-07-20 04:58:24
虽然这当然可以在base R中完成，但Hadley Wickham的工具提供了一种很好的方法来实现这一点(通过某种标准拆分数据帧，对子集应用转换，然后将它们重新粘合在一起-简称为“拆分-应用-组合”)。
(这些都是未经测试的，但我认为它们应该可以工作...)
library(plyr)
library(dplyr) ## load dplyr *second!*plyr (Hadleyverse/tidyverse 1)
ddply(my_data,"Experiment",
mutate,
Correction.Factor=Loading.Control/
Loading.Control[Treatment=="None" & Inhibitor=="DMSO"])dplyr (Hadleyverse/tidyverse 2)
my_data %>% group_by(Experiment) %>%
mutate(Correction.Factor=Loading.Control/
Loading.Control[Treatment=="None" & Inhibitor=="DMSO"])另一种(base-R友好的)方法是细分基线(None/DMSO)情况，只保留实验和Loading.Control列(但重命名后者)，然后用原始数据集merge返回结果-然后您所要做的就是将一列除以另一列……
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/38468285
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