Question

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我很擅长在R中编写循环或函数，但我仍然没有真正理解如何做到这一点。目前，我需要编写一个循环/函数（不确定哪一个会更好）来执行几个不同数据帧的t测试。

我的数据类似于：

set.seed(694)
df_1_08 <- data.frame(
  year = 2008,
  a = runif(100, 0, 100),
  b = runif(100, 0, 100),
  c = runif(100, 0, 100),
  d = runif(100, 0, 100)
)

df_1_09 <- data.frame(
  year = 2009,
  a = rnorm(100, 0, 1),
  b = rnorm(100, 0, 1),
  c = rnorm(100, 0, 1),
  d = rnorm(100, 0, 1)
)


df_1_10 <- data.frame(
  year = 2010,
  a = rnorm(100, 0, 1),
  b = rnorm(100, 0, 1),
  c = rnorm(100, 0, 1),
  d = rnorm(100, 0, 1)
)


df_2_08 <- data.frame(
  year = 2008,
  a = rnorm(100, 0, 1),
  b = rnorm(100, 0, 1),
  c = rnorm(100, 0, 1),
  d = rnorm(100, 0, 1)
)


df_2_09 <- data.frame(
  year = 2009,
  a = rnorm(100, 0, 1),
  b = rnorm(100, 0, 1),
  c = rnorm(100, 0, 1),
  d = rnorm(100, 0, 1)
)


df_2_10 <- data.frame(
  year = 2010,
  a = rnorm(100, 0, 1),
  b = rnorm(100, 0, 1),
  c = rnorm(100, 0, 1),
  d = rnorm(100, 0, 1)
)

# Write Loop to do t-test between dfs 08, 09, 10 comparing columns a, b, c, d and storing the full results in a df

基本上，我需要对此数据执行的操作是每年对特定列运行t检验（2008,2009,2010），以便df_1_08运行df_2_08的t检验所有列（a，b，c，d）然后将这些列存储在数据框中（其中存储了t统计量，p值等））。这听起来像是一个完美的循环工作。但是我也需要每年（2008年，2009年和2010年）都这样做，并将结果存储在不同的数据框中，所以这听起来像是一个完美的功能。

我不确定如何写，所以我想我在编写这些循环/函数时要求一些帮助。提前感谢您提供的任何帮助。

我还可以将数据帧组合成一个大的df，其中一列标识原始数据帧号（即df1或df2），一列标识数据帧年（即2008,2009,2010）。它看起来像这样：

df1 <- rbind(df_1_08, df_1_09, df_1_10)
df1$ID <-1
df2 <- rbind(df_1_08, df_1_09, df_1_10)
df2$ID <- 2

master.df <- rbind(df1, df2)

我不确定编写一个循环/函数来运行带有master.df的t.tests是否更容易。在那个df中，我基本上需要在循环或函数中执行以下操作：

将master.df子集设为df1和df2
年内的子集df1和df2
每年为t.test，a，b和c列运行d
将所有相关的t.test输出（即t统计，p值等）存储在我可以打印的data.frame中。

Answer 1

怎么样：

df_1_08 <- data.frame(year = 2008, a = runif(100, 0, 100), b = runif(100, 0, 100), c = runif(100, 0, 100), d = runif(100, 0, 100))
df_1_09 <- data.frame(year = 2009, a = runif(100, 0, 100), b = runif(100, 0, 100), c = runif(100, 0, 100), d = runif(100, 0, 100))
df_1_10 <- data.frame(year = 2010, a = runif(100, 0, 100), b = runif(100, 0, 100), c = runif(100, 0, 100), d = runif(100, 0, 100))
df_2_08 <- data.frame(year = 2008, a = runif(100, 0, 100), b = runif(100, 0, 100), c = runif(100, 0, 100), d = runif(100, 0, 100))
df_2_09 <- data.frame(year = 2009, a = runif(100, 0, 100), b = runif(100, 0, 100), c = runif(100, 0, 100), d = runif(100, 0, 100))
df_2_10 <- data.frame(year = 2010, a = runif(100, 0, 100), b = runif(100, 0, 100), c = runif(100, 0, 100), d = runif(100, 0, 100))

dfs_1.names <- ls()[grep("df_1", ls())]
dfs_2.names <- ls()[grep("df_2", ls())]
dfs_1.list <-lapply(dfs_1.names, get)
dfs_2.list <- lapply(dfs_2.names, get)

#in case you want to try the matrix
dfs_1.mtrx <- do.call("rbind",dfs_1.list)
dfs_2.mtrx <- do.call("rbind",dfs_2.list)

years <- intersect(unique(dfs_1.mtrx[,"year"]),unique(dfs_2.mtrx[,"year"]))
# [1] 2008 2009 2010
columns <- intersect(colnames(dfs_1.mtrx[,-1]),colnames(dfs_2.mtrx[,-1]))
# [1] "a" "b" "c" "d"

df.ttest <- as.data.frame(matrix(NA, ncol = 8, nrow = length(years)*length(columns)))
colnames(df.ttest) <- c("year","column","tstat","p.value","degreesf","low.conf","up.conf","data.name")
count = 0
for(i in 1:length(years)){
  for(j in columns){
    ttest <- t.test(dfs_1.list[[i]][j], dfs_2.list[[i]][j])
    ttest$data.name <- paste(paste0("df_1_",years[i]-2000,"$",j),"and",
                             paste0("df_2_",years[i]-2000,"$",j))
    count <- count + 1
    df.ttest[count, "year"]     <- years[i]
    df.ttest[count, "column"]   <- j
    df.ttest[count, "tstat"]    <- ttest$statistic
    df.ttest[count, "p.value"]  <- ttest$p.value
    df.ttest[count, "degreesf"] <- ttest$parameter
    df.ttest[count, "low.conf"] <- ttest$conf.int[1]
    df.ttest[count, "up.conf"]  <- ttest$conf.int[2]   
    df.ttest[count, "data.name"] <- ttest$data.name
  }
}
df.ttest

看起来像：

   year column      tstat    p.value degreesf    low.conf   up.conf               data.name
1  2008      a  1.0607688 0.29008725 197.9914  -3.7038792 12.327117   df_1_8$a and df_2_8$a
2  2008      b  0.3311722 0.74086573 197.3689  -6.6956039  9.398291   df_1_8$b and df_2_8$b
3  2008      c  1.0410813 0.29910773 197.9405  -3.7582835 12.164152   df_1_8$c and df_2_8$c
4  2008      d  1.2623350 0.20834791 193.4532  -2.9384999 13.387911   df_1_8$d and df_2_8$d
5  2009      a -0.5764091 0.56500626 194.1686 -10.1442158  5.555762   df_1_9$a and df_2_9$a
6  2009      b -1.5222524 0.12954190 197.9248 -14.4317793  1.857603   df_1_9$b and df_2_9$b
7  2009      c -0.1744245 0.86171283 195.0217  -8.6590932  7.251902   df_1_9$c and df_2_9$c
8  2009      d  0.0839337 0.93319409 197.6654  -7.5768817  8.250526   df_1_9$d and df_2_9$d
9  2010      a  1.9125742 0.05724768 197.7406  -0.2353887 15.378495 df_1_10$a and df_2_10$a
10 2010      b  0.9024489 0.36792603 196.0224  -4.0977460 11.011904 df_1_10$b and df_2_10$b
11 2010      c -0.9735756 0.33145768 197.5899 -12.2641333  4.157135 df_1_10$c and df_2_10$c
12 2010      d  0.8721498 0.38418378 197.8601  -4.5311820 11.717207 df_1_10$d and df_2_10$d

为各种t检验编写循环或函数

1 个答案: