我是R的新手，想用它来让我的生活更容易分析我的荧光检测数据。 在我在excel中手动进行分析之前，现在想通过为它设置r脚本来简化它。
我的数据的一个例子是：
> df <- data.frame(time=1:10, sample1=4*1:10, sample2=3*1:10, sample3=2*1:10)
> df
time sample1 sample2 sample3
1 1 4 3 2
2 2 8 6 4
3 3 12 9 6
4 4 16 12 8
5 5 20 15 10
6 6 24 18 12
7 7 28 21 14
8 8 32 24 16
9 9 36 27 18
10 10 40 30 20
所以第一列始终是我的分析时间，下面的列代表每个样品在给定时间点的荧光信号。
然后，我想计算每个样品的荧光随时间的斜率（例如样品1随时间变化，样品2随时间变化，样品3随时间变化，......）。结果我应该得到每列的一个斜率值。在Excel中我用过：斜率（B2：BX; $ A $ 2：$ A $ X） 由于我通常有96个样本，这使得在excel中手动执行它更加烦人。
@missuse提供的解决方案
apply(df[,2:ncol(df)], 2, function(x){
model = lm(x ~ df$time - 1)
return(coef(model))
})
为我上面展示的样本工作，但不是我的真实数据。
根据要求，这是我的真实数据的一部分：
> df1
Time 1 2 3 4 5 6 7 8 9 10 11 12 13
1 0 24315.5 21446.5 46748.5 36008 15501 16799.5 24847 25354.5 16617.5 10576 43422.5 40036 15988.5
2 26.1 25592.5 22667.5 47310.0 36284.0 15790.5 16815.5 25108.0 25535.0 16702.5 10418.0 43602.0 40301.5 16227.0
3 52.1 26493.0 22839.5 47356.5 36549.0 15773.5 16804.0 25307.5 25538.5 16697.5 10390.0 43682.0 40332.0 16271.0
4 78.2 26889 23585 47496.5 36525 15942.5 16903 25498 25565 16796.5 10369.5 43768.5 40253.5 16584
5 104.3 27320.5 23914 47331.5 36680.5 16033.5 16912 25717 25798.5 16903.5 10356.5 43960 40299 16604.5
6 130.4 27823.0 24208.0 47815.0 36591.0 16132.0 17052.5 25669.5 25614.5 16958.0 10306.0 44104.5 40266.0 16682.5
7 156.4 28335.0 24647.0 47838.0 36718.0 16269.5 17001.0 25945.0 25754.5 16995.0 10397.5 43998.5 40256.5 16838.5
8 182.5 28859.5 25128 48056 36887 16385 17032.5 25832.5 25710 16980.5 10306.5 44282.5 40461 16995
9 208.6 29324.0 25369.0 48094.5 36889.5 16360.5 16931.0 25961.0 25918.5 17259.0 10271.0 44297.0 40511.0 17033.0
10 234.6 29920.5 25803.5 48314.5 36755.5 16566.0 17106.5 26210.0 25595.5 17293.5 10268.5 44355.5 40503.5 17047.5
11 260.7 30412.5 26314.5 48709.5 36848.5 16630.5 17208.0 26065.0 25702.0 17448.5 10296.0 44805.0 40383.5 17029.5
12 286.8 30883.0 26624.0 48570.5 36845.0 16804.0 17116.5 26237.0 25817.0 17523.0 10274.0 44743.5 40727.5 17167.5
> dput(df1[1:5, 1:3])
structure(list(Time = structure(c(44L, 1L, 2L, 47L, 45L), .Names = c("X__2",
"X__3", "X__4", "X__5", "X__6"), .Label = c(" 26.1", " 52.1",
" 130.4", " 156.4", " 208.6", " 234.6", " 260.7", " 286.8",
" 312.8", " 338.9", " 365.0", " 391.0", " 417.1", " 443.2",
" 469.2", " 495.3", " 521.4", " 547.4", " 573.5", " 599.6",
" 625.6", " 651.7", " 677.8", " 703.9", " 729.9", " 756.0",
" 782.1", " 808.1", " 834.2", " 860.3", " 886.3", " 912.4",
" 938.5", " 964.5", " 990.6", " 1016.7", " 1042.7", " 1068.8",
" 1094.9", " 1120.9", " 1147.0", " 1173.1", " 1199.1", "0", "104.3",
"182.5", "78.2"), class = "factor"), `1` = structure(1:5, .Names = c("X__2",
"X__3", "X__4", "X__5", "X__6"), .Label = c("24315.5", "25592.5",
"26493.0", "26889", "27320.5", "27823.0", "28335.0", "28859.5",
"29324.0", "29920.5", "30412.5", "30883.0", "31599.5", "31958.0",
"32744.0", "33065.5", "33432.0", "34269.0", "34603.5", "35214.5",
"35570.5", "36149.0", "36596.5", "37087.5", "37520.0", "38254.5",
"38540.5", "39200.5", "39718.0", "40126.0", "40808.0", "41235.0",
"41537.5", "42316.5", "42755.5", "42927.0", "43772.0", "44095.0",
"44669.0", "45027.0", "45607.0", "45976.5", "46624.5", "46961.0",
"47338.0", "48147.5", "48499.0"), class = "factor"), `2` = structure(1:5, .Names = c("X__2",
"X__3", "X__4", "X__5", "X__6"), .Label = c("21446.5", "22667.5",
"22839.5", "23585", "23914", "24208.0", "24647.0", "25128", "25369.0",
"25803.5", "26314.5", "26624.0", "27103.5", "27366.5", "27656.5",
"28195.0", "28655.0", "28912.5", "29316.5", "29530.0", "29931.0",
"30401.5", "30899.0", "30973.5", "31643.0", "31740.5", "32313.0",
"32597.5", "32967.0", "33331.5", "33825.0", "34051.5", "34438.0",
"34646.0", "35299.0", "35365.5", "35980.0", "36217.0", "36634.0",
"37005.0", "37338.5", "37842.0", "38039.0", "38501.5", "38694.0",
"39057.5", "39330.5"), class = "factor")), .Names = c("Time",
"1", "2"), row.names = c(NA, 5L), class = "data.frame")
当我使用@missuse提供的解决方案时，我得到以下内容而不是每列的单个斜率值：
> df1
1 2 3 4 5 6 7 8 9 10 11 12 13
data8$Time 26.1 25592.5 22667.5 47310.0 36284.0 15790.5 16815.5 25108.0 25535.0 16702.5 10418.0 43602.0 40301.5 16227.0
data8$Time 52.1 26493.0 22839.5 47356.5 36549.0 15773.5 16804.0 25307.5 25538.5 16697.5 10390.0 43682.0 40332.0 16271.0
data8$Time 130.4 27823.0 24208.0 47815.0 36591.0 16132.0 17052.5 25669.5 25614.5 16958.0 10306.0 44104.5 40266.0 16682.5
data8$Time 156.4 28335.0 24647.0 47838.0 36718.0 16269.5 17001.0 25945.0 25754.5 16995.0 10397.5 43998.5 40256.5 16838.5
data8$Time 208.6 29324.0 25369.0 48094.5 36889.5 16360.5 16931.0 25961.0 25918.5 17259.0 10271.0 44297.0 40511.0 17033.0
data8$Time 234.6 29920.5 25803.5 48314.5 36755.5 16566.0 17106.5 26210.0 25595.5 17293.5 10268.5 44355.5 40503.5 17047.5
data8$Time 260.7 30412.5 26314.5 48709.5 36848.5 16630.5 17208.0 26065.0 25702.0 17448.5 10296.0 44805.0 40383.5 17029.5
data8$Time 286.8 30883.0 26624.0 48570.5 36845.0 16804.0 17116.5 26237.0 25817.0 17523.0 10274.0 44743.5 40727.5 17167.5
data8$Time 312.8 31599.5 27103.5 48943.5 36966.0 16807.0 17150.0 26306.5 25697.0 17566.0 10375.0 44674.0 40740.5 17309.0
答案 0 :(得分：1)
这是一个可能的解决方案
apply(df[,2:ncol(df)], 2, function(x){
model = lm(x ~ df$time - 1)
return(coef(model))
})
#ouput
sample1 sample2 sample3
4 3 2
假设截距为0.如果您希望使用截距评估，请使用：
model = lm(x ~ df$time)
此代码的作用是apply数据框（2）的列（由df表示）上的函数。它从第二列到最后一列（[2:ncol(df]）获取所有列，并进行线性回归（model = lm(x ~ df$time - 1）并返回系数（return(coef(model)）。
如果第一列的名称并不总是time，那么：
model = lm(x ~ df[,1] - 1)
表示第一列是x
编辑：问题是变量未编码为数字。这是一个解决方案：
df$Time = as.numeric(df$Time) #convert time to numeric
没有拦截：
apply(df[,2:ncol(df)], 2, function(x){
x = as.numeric(x) #convert x to numeric
model = lm(x ~ df$Time - 1)
return(coef(model))
})
拦截：
apply(df[,2:ncol(df)], 2, function(x){
x = as.numeric(x) #convert x to numeric
model = lm(x ~ df$time)
return(coef(model)[2])
})