使用R每天随机分配参与者参加治疗 [英] Use R to Randomly Assign of Participants to Treatments on a Daily Basis
问题描述
问题:
我正在尝试使用R生成随机研究设计，其中一半的参与者被随机分配给治疗1"，另一半的参与者被分配给治疗2".但是，由于一半的受试者是男性，一半的受试者是女性，我还想确保每次治疗都接受相同数量的男性和女性，因此，应该将一半的男性和女性分配给治疗1"，其余的一半应分配给处理2".
I am attempting to use R to generate a random study design where half of the participants are randomly assigned to "Treatement 1" and the other half are assigned to "Treatment 2". However, because half of the subjects are male and half are female and I also want to ensure that an equal number of males and females are exposed to each treatment, half of the males and females should be assigned to "Treatment 1" and the remaining half should be assigned to "Treatment 2".
此设计有两个并发症:(1)这项为期一年的研究，必须每天对参与者进行治疗； (2)每位参与者必须在28天内至少接触10次治疗1".
There are two complications to this design: (1) This is a yearlong study and the assignment of participants to treatment must occur on a daily basis; and (2) Each participant must be exposed to "Treatment 1" a minimum 10 times in a 28 day period.
这是否有可能在R界面中将其自动化?我以为是这样，但是我认为我作为R程序员的初学者身份使我无法自己寻找解决方案.我已经花了几天的时间来弄清楚如何实现它，并且浏览了该站点上许多听起来相像的帖子，但这些帖子在这里未能成功应用.我希望外面的人知道一些技巧，可以帮助我解决这个问题，任何建议将不胜感激！
Is this even possible to automate this in the R interface? I assume so, but I think my beginner status as an R programmer prohibits me from finding the solution on my own. I have been struggling for days to figure out how to actualize this, and have looked through many similar-sounding posts on this site that were not able to be successfully applied here. I am hoping someone out there knows some tricks that could help me get unstuck in solving this problem, any advice would be greatly appreciated!
我尝试过的事情:
特定信息
# There are 16 participants
p <- c("P01", "P02", "P03", "P04", "P05", "P06", "P07", "P08", "P09", "P10", "P11", "P12", "P13", "P14", "P15", "P16")
# Half are male and half are female
g <- c(rep("M", 8), rep("F", 8))
# I make a dataframe but this may not be necessary
df <- cbind.data.frame(p,g)
# There are 365 days in one year
d <- seq(1,365,1)
...不幸的是，我不确定如何从这里继续.
...unfortunately, I am not sure how to proceed from here.
理想结果:
我正在预想与该表近似的结果:
I am envisioning something approximate to this table as the outcome:
基本上，每个参与者都有一列，每天都有一行.与每天相关联的是治疗1(T1)或治疗2(T2)的分配，其中8位男性中的4位和8位女性中的4位被分配给T1，其余部分分配给T2.每天将这些治疗重新分配1年.此图表中未描述每个参与者在28天的时间里至少暴露10次T1的需求.如果还有其他意义，该表不必看起来像这样！
Basically there is a column for each participant and a row for each day. Associated with each day is an assignment to either Treatment 1 (T1) or Treatment 2 (T2), with 4 of the 8 males and 4 of the 8 females being assigned to T1 and the remainder to T2. These treatments are reassigned every day for 1 year. Not depicted in this chart is the need for each participant to be exposed to T1 at least 10 times in a 28-day period. The table does not have to look like that if something else makes more sense!
推荐答案
考虑使用by按 day 和 gender 划分数据帧，然后使用by运行足够的样本c1>在100次选择平衡治疗的几种方法之一:
Consider splitting data frame by day and gender with by, then run enough samples with replicate at 100 times to pick one of several where treatments are balanced:
数据
df <- merge(data.frame(participant = p, gender = g),
data.frame(days = seq(1,365)),
by=NULL)
解决方案
df_list <- by(df, list(df$gender, df$days), function(sub){
t <- replicate(100, { # RUN 100 REPETITIONS OF EXPRESSION
s <- sample(c("T1", "T2"), size=nrow(sub), replace=TRUE) # SAMPLE "T1" AND "T2" BY SIZE OF SUBSET
s[ sum(s == "T1") == sum(s == "T2") ] # FILTER TO EQUAL TREATMENTS
})
t <- Filter(length, t)[[1]] # SELECT FIRST OF SEVERAL NON-EMPTY RETURNS
transform(sub, treatment = t) # ASSIGN RESULT TO NEW COLUMN
})
# BIND DATA FRAMES AND RESET ROW.NAMES
final_df <- data.frame(do.call(rbind.data.frame, df_list), row.names=NULL)
输出
第1天
head(final_df, 16)
# participant gender days treatment
# 1 P09 F 1 T1
# 2 P10 F 1 T2
# 3 P11 F 1 T2
# 4 P12 F 1 T1
# 5 P13 F 1 T2
# 6 P14 F 1 T2
# 7 P15 F 1 T1
# 8 P16 F 1 T1
# 9 P01 M 1 T1
# 10 P02 M 1 T1
# 11 P03 M 1 T2
# 12 P04 M 1 T2
# 13 P05 M 1 T2
# 14 P06 M 1 T1
# 15 P07 M 1 T1
# 16 P08 M 1 T2
365天
tail(final_df, 16)
# participant gender days treatment
# 5825 P09 F 365 T2
# 5826 P10 F 365 T2
# 5827 P11 F 365 T1
# 5828 P12 F 365 T2
# 5829 P13 F 365 T1
# 5830 P14 F 365 T2
# 5831 P15 F 365 T1
# 5832 P16 F 365 T1
# 5833 P01 M 365 T1
# 5834 P02 M 365 T2
# 5835 P03 M 365 T1
# 5836 P04 M 365 T2
# 5837 P05 M 365 T2
# 5838 P06 M 365 T2
# 5839 P07 M 365 T1
# 5840 P08 M 365 T1
理想地，出于分析目的，您应将数据保留为长格式(即整洁数据).但是，如果需要宽格式，请考虑使用reshape进行帮助和清除处理:
Ideally, for analytical purposes you should keep data in long format (i.e., tidy data). But if needing wide format consider reshape with helper and cleanup processing:
# HELPER OBJECTS
final_df$participant_gender <- with(final_df, paste0(participant, gender))
new_names <- paste0(p, g)
# RESHAPE WIDE
wide_df <- reshape(final_df, v.names = "treatment", timevar = "participant_gender",
idvar="days", drop = c("gender", "participant"),
new.row.names = 1:365, direction = "wide")
# RENAME AND RE-ORDER COLUMNS
names(wide_df) <- gsub("treatment.", "", names(wide_df))
wide_df <- wide_df[c("days", new_names)]
head(wide_df)
# days P01M P02M P03M P04M P05M P06M P07M P08M P09F P10F P11F P12F P13F P14F P15F P16F
# 1 1 T1 T1 T2 T2 T2 T1 T1 T2 T1 T2 T2 T1 T2 T2 T1 T1
# 2 2 T1 T1 T2 T1 T2 T1 T2 T2 T1 T2 T2 T1 T2 T2 T1 T1
# 3 3 T1 T1 T2 T1 T1 T2 T2 T2 T1 T2 T2 T2 T1 T2 T1 T1
# 4 4 T1 T1 T1 T2 T2 T2 T1 T2 T2 T1 T1 T2 T2 T1 T1 T2
# 5 5 T1 T1 T2 T1 T2 T2 T1 T2 T1 T1 T2 T1 T2 T2 T1 T2
# 6 6 T2 T1 T1 T1 T2 T2 T1 T2 T2 T2 T2 T1 T2 T1 T1 T1
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