对具有许多列的两个数据框列表执行多个两个样本 t 检验
Performing multiple two sample t-tests on two lists of data frames that have many columns
我有两个列表,每个列表有四个数据框。两个列表(“loc_list_future”和“loc_list_2019”中的数据框都有 33 列:“年”,然后是 32 种不同气候模型的平均降水值。
loc_list_future 中的数据框看起来像这样,但总共有 32 个模型列,数据转到 2059 年:
Year Model 1 Model 2 Model 3 ...Model 32
2020 714.1101 686.5888 1048.4274
2021 1018.0095 766.9161 514.2700
2022 756.7066 902.2542 906.2877
2023 906.9675 919.5234 647.6630
2024 767.4008 861.1275 700.2612
2025 876.1538 738.8370 664.3342
2026 781.5092 801.2387 743.8965
2027 876.3522 819.4323 675.3022
2028 626.9468 927.0774 696.1884
2029 752.4084 824.7682 835.1566
...
2059
loc_list_2019 中的数据框的年份范围为 2006-2019,但其他方面看起来相同。
每个数据框代表一个地理位置,两个列表具有相同的四个位置,但一个列表用于 2006-2019 年的值,另一个用于未来的值。
我想要 运行 两个样本 t 检验,将 2006-19 年的值与每个模型在每个位置的未来值进行比较。
我有另一个列表 (loc_list_OBS),它的数据框只有两列“Year”和“Mean_Precip”(这是观察到的数据,不是基于模型,这就是为什么只有一列表示平均降水量)。我的代码(见下文)将针对未来数据(loc_list_future)对观察到的数据(loc_list_OBS)进行 运行 双样本 t 检验,但我不确定如何才能将此代码更改为 运行 两个列表的 t 检验,每个列表有 32 个模型。
myfun <- function(x,y)
{
OBS_Data <- x$Mean_Precip
#Empty list
List <- list()
#Now loop
for(i in 2:dim(y)[2])
{
#Label
val <- names(y[,i,drop=F])
Future_Data <- y[,i]
#Test
test <- t.test(OBS_Data, Future_Data, alternative = "two.sided")
#Save
List[[i-1]] <- test
names(List)[i-1] <- val
}
return(List)
}
t.stat <- mapply(FUN = myfun,x=loc_list_OBS,y=loc_list_future, SIMPLIFY = FALSE)
我建议采用下一种方法。我创建了类似于您所拥有的虚拟数据。这里的代码:
#Data before
dfb <- structure(list(Year = 2010:2019, Model.1 = c(614.1101, 918.0095,
656.7066, 806.9675, 667.4008, 776.1538, 681.5092, 776.3522, 526.9468,
652.4084), Model.2 = c(586.5888, 666.9161, 802.2542, 819.5234,
761.1275, 638.837, 701.2387, 719.4323, 827.0774, 724.7682), Model.3 = c(948.4274,
414.27, 806.2877, 547.663, 600.2612, 564.3342, 643.8965, 575.3022,
596.1884, 735.1566)), class = "data.frame", row.names = c(NA,
-10L))
#Data after
dfa <- structure(list(Year = 2020:2029, Model.1 = c(714.1101, 1018.0095,
756.7066, 906.9675, 767.4008, 876.1538, 781.5092, 876.3522, 626.9468,
752.4084), Model.2 = c(686.5888, 766.9161, 902.2542, 919.5234,
861.1275, 738.837, 801.2387, 819.4323, 927.0774, 824.7682), Model.3 = c(1048.4274,
514.27, 906.2877, 647.663, 700.2612, 664.3342, 743.8965, 675.3022,
696.1884, 835.1566)), class = "data.frame", row.names = c(NA,
-10L))
现在代码:
#Data for lists
L.before <- list(df1=dfb,df2=dfb,df3=dfb,df4=dfb)
L.after <- list(df1=dfa,df2=dfa,df3=dfa,df4=dfa)
函数:
#Function
myfun <- function(x,y)
{
#Create empty list
List <- list()
#Loop
for(i in 2:dim(x)[2])
{
name <- names(x[,i,drop=F])
before <- x[,i]
after <- y[,i]
#Test
test <- t.test(before, after, alternative = "two.sided")
#Save
List[[i-1]] <- test
names(List)[i-1] <- name
}
return(List)
}
申请:
#Apply
t.stat <- mapply(FUN = myfun,x=L.before,y=L.after, SIMPLIFY = FALSE)
一些输出:
t.stat[[1]]
$Model.1
Welch Two Sample t-test
data: before and after
t = -1.9966, df = 18, p-value = 0.06122
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-205.224021 5.224021
sample estimates:
mean of x mean of y
707.6565 807.6565
$Model.2
Welch Two Sample t-test
data: before and after
t = -2.8054, df = 18, p-value = 0.0117
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-174.88934 -25.11066
sample estimates:
mean of x mean of y
724.7764 824.7764
$Model.3
Welch Two Sample t-test
data: before and after
t = -1.4829, df = 18, p-value = 0.1554
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-241.67613 41.67613
sample estimates:
mean of x mean of y
643.1787 743.1787
让我知道这是否适合您!
我有两个列表,每个列表有四个数据框。两个列表(“loc_list_future”和“loc_list_2019”中的数据框都有 33 列:“年”,然后是 32 种不同气候模型的平均降水值。
loc_list_future 中的数据框看起来像这样,但总共有 32 个模型列,数据转到 2059 年:
Year Model 1 Model 2 Model 3 ...Model 32
2020 714.1101 686.5888 1048.4274
2021 1018.0095 766.9161 514.2700
2022 756.7066 902.2542 906.2877
2023 906.9675 919.5234 647.6630
2024 767.4008 861.1275 700.2612
2025 876.1538 738.8370 664.3342
2026 781.5092 801.2387 743.8965
2027 876.3522 819.4323 675.3022
2028 626.9468 927.0774 696.1884
2029 752.4084 824.7682 835.1566
...
2059
loc_list_2019 中的数据框的年份范围为 2006-2019,但其他方面看起来相同。
每个数据框代表一个地理位置,两个列表具有相同的四个位置,但一个列表用于 2006-2019 年的值,另一个用于未来的值。
我想要 运行 两个样本 t 检验,将 2006-19 年的值与每个模型在每个位置的未来值进行比较。
我有另一个列表 (loc_list_OBS),它的数据框只有两列“Year”和“Mean_Precip”(这是观察到的数据,不是基于模型,这就是为什么只有一列表示平均降水量)。我的代码(见下文)将针对未来数据(loc_list_future)对观察到的数据(loc_list_OBS)进行 运行 双样本 t 检验,但我不确定如何才能将此代码更改为 运行 两个列表的 t 检验,每个列表有 32 个模型。
myfun <- function(x,y)
{
OBS_Data <- x$Mean_Precip
#Empty list
List <- list()
#Now loop
for(i in 2:dim(y)[2])
{
#Label
val <- names(y[,i,drop=F])
Future_Data <- y[,i]
#Test
test <- t.test(OBS_Data, Future_Data, alternative = "two.sided")
#Save
List[[i-1]] <- test
names(List)[i-1] <- val
}
return(List)
}
t.stat <- mapply(FUN = myfun,x=loc_list_OBS,y=loc_list_future, SIMPLIFY = FALSE)
我建议采用下一种方法。我创建了类似于您所拥有的虚拟数据。这里的代码:
#Data before
dfb <- structure(list(Year = 2010:2019, Model.1 = c(614.1101, 918.0095,
656.7066, 806.9675, 667.4008, 776.1538, 681.5092, 776.3522, 526.9468,
652.4084), Model.2 = c(586.5888, 666.9161, 802.2542, 819.5234,
761.1275, 638.837, 701.2387, 719.4323, 827.0774, 724.7682), Model.3 = c(948.4274,
414.27, 806.2877, 547.663, 600.2612, 564.3342, 643.8965, 575.3022,
596.1884, 735.1566)), class = "data.frame", row.names = c(NA,
-10L))
#Data after
dfa <- structure(list(Year = 2020:2029, Model.1 = c(714.1101, 1018.0095,
756.7066, 906.9675, 767.4008, 876.1538, 781.5092, 876.3522, 626.9468,
752.4084), Model.2 = c(686.5888, 766.9161, 902.2542, 919.5234,
861.1275, 738.837, 801.2387, 819.4323, 927.0774, 824.7682), Model.3 = c(1048.4274,
514.27, 906.2877, 647.663, 700.2612, 664.3342, 743.8965, 675.3022,
696.1884, 835.1566)), class = "data.frame", row.names = c(NA,
-10L))
现在代码:
#Data for lists
L.before <- list(df1=dfb,df2=dfb,df3=dfb,df4=dfb)
L.after <- list(df1=dfa,df2=dfa,df3=dfa,df4=dfa)
函数:
#Function
myfun <- function(x,y)
{
#Create empty list
List <- list()
#Loop
for(i in 2:dim(x)[2])
{
name <- names(x[,i,drop=F])
before <- x[,i]
after <- y[,i]
#Test
test <- t.test(before, after, alternative = "two.sided")
#Save
List[[i-1]] <- test
names(List)[i-1] <- name
}
return(List)
}
申请:
#Apply
t.stat <- mapply(FUN = myfun,x=L.before,y=L.after, SIMPLIFY = FALSE)
一些输出:
t.stat[[1]]
$Model.1
Welch Two Sample t-test
data: before and after
t = -1.9966, df = 18, p-value = 0.06122
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-205.224021 5.224021
sample estimates:
mean of x mean of y
707.6565 807.6565
$Model.2
Welch Two Sample t-test
data: before and after
t = -2.8054, df = 18, p-value = 0.0117
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-174.88934 -25.11066
sample estimates:
mean of x mean of y
724.7764 824.7764
$Model.3
Welch Two Sample t-test
data: before and after
t = -1.4829, df = 18, p-value = 0.1554
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-241.67613 41.67613
sample estimates:
mean of x mean of y
643.1787 743.1787
让我知道这是否适合您!