计算荧光随时间的斜率 [英] Calculating slopes of fluorescence over time
本文介绍了计算荧光随时间的斜率的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我是R的新手，并希望使用它使我的生活更轻松地分析我的荧光测定数据.在我之前使用excel手动进行分析之前，现在想通过设置R脚本来简化它.
我的数据的一个例子是这样:
>df<-data.frame(时间= 1:10，sample1 = 4 * 1:10，sample2 = 3 * 1:10，sample3 = 2 * 1:10)>df时间采样1采样2采样31 1 4 3 22 2 8 6 43 3 12 9 64 4 16 12 85 5 20 15 106 6 24 18 127 7 28 21 148 8 32 24 169 9 36 27 1810 10 40 30 20 因此，第一列始终是我的测定时间，接下来的列代表给定时间点上每个样品的荧光信号.
然后我要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，我应该获得每列一个斜率值.在excel中，我使用了:坡度(B2:BX; $ A $ 2:$ A $ X)由于我通常有96个样本，因此用excel手动处理会更加烦人.
@missuse提供的解决方案
apply(df [，2:ncol(df)]，2，function(x){模型= lm(x〜df $ time-1)返回(coef(model))}) 适用于上面显示的示例df，但不适用于我的真实数据.
根据要求，这是我的真实数据的一部分:
>df1时间1 2 3 4 5 6 7 8 9 10 11 12 131 0 24315.5 21446.5 46748.5 36008 15501 16799.5 24847 25354.5 16617.5 10576 43422.5 40036 15988.52 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 16227.03 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 16271.04 78.2 26889 23585 47496.5 36525 15942.5 16903 25498 25565 16796.5 10369.5 43768.5 40253.5 165845 104.3 27320.5 23914 47331.5 36680.5 16033.5 16912 25717 25798.5 16903.5 10356.5 43960 40299 16604.56 130.4 27823.0 24208.0 47815.0 36591.0 16132.0 17052.5 25669.5 25614.5 16958.0 10306.0 44104.5 40266.0 16682.57 156.4 28335.0 24647.0 47838.0 36718.0 16269.5 17001.0 25945.0 25754.5 16995.0 10397.5 43998.5 40256.5 16838.58 182.5 28859.5 25128 48056 36887 16385 17032.5 25832.5 25710 16980.5 10306.5 44282.5 40461 169959 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 17033.010 234.6 29920.5 25803.5 48314.5 36755.5 16566.0 17106.5 26210.0 25595.5 17293.5 10268.5 44355.5 40503.5 17047.511 260.7 30412.5 26314.5 48709.5 36848.5 16630.5 17208.0 26065.0 25702.0 17448.5 10296.0 44805.0 40383.5 17029.512 286.8 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")，类="factor")，`2` =结构(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，我得到的是以下内容，而不是每一列的单个斜率值:
>df11 2 3 4 5 6 7 8 9 10 11 12 13data8 $ 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 16227.0data8 $ 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 16271.0data8 $ 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 16682.5data8 $ 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 16838.5data8 $ 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 17033.0data8 $ 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 17047.5data8 $ 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 17029.5data8 $ 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 17167.5data8 $ 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 解决方案
这是一个可能的解决方案
apply(df [，2:ncol(df)]，2，function(x){模型= lm(x〜df $ time-1)返回(coef(model))})#输出样品1样品2样品34 3 2 假定截距为0.如果您希望对截距进行评估，请使用:
模型= lm(x〜df $ time)
此代码的作用是在数据框( df )的列(由 2 表示)上应用一个函数.它从第二个到最后一个取所有列( [2:ncol(df] ))，并进行线性回归( model = lm(x〜df $ time-1 .并返回系数( return(coef(模型)).
如果第一列并非始终命名为 time ，则:
模型= lm(x〜df [，1]-1)
表示第一列是 x
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)#将时间转换为数值 没有拦截:
apply(df [，2:ncol(df)]，2，function(x){x = as.numeric(x)#将x转换为数值模型= lm(x〜df $ Time-1)返回(coef(model))}) 使用拦截:
apply(df [，2:ncol(df)]，2，function(x){x = as.numeric(x)#将x转换为数值模型= lm(x〜df $ time)return(coef(model)[2])}) I am a newbie to R and want to use it to make my life easier analyzing the data of my fluorescence assays. Before I did the analysis manually in excel but now want to make it easier by setting up a r-script for it.
one example of my data would be this:
> 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
so the first column is always the time of my assay and the following columns represent the fluorescent signal of each sample for a given time-point.
I then want to calculate the slope of fluorescence over time for every sample (e.g. sample1 over time, sample2 over time, sample3 over time,...). As a result I should get one value of slope per column.
In excel I used: slope(B2:BX;$A$2:$A$X) Since I usually have 96 samples this makes it even more annoying to do it by hand in excel.
The solution offered by @missuse
apply(df[,2:ncol(df)], 2, function(x){
model = lm(x ~ df$time - 1)
return(coef(model))
})
worked for the sample df I showed above but not for my real data.
As requested, here is a part of my real data:
> 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")
When I use the solution offered by @missuse I get the following instead of a single slope value for each column:
> 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$Time 312.8 31599.5 27103.5 48943.5 36966.0 16807.0 17150.0 26306.5 25697.0 17566.0 10375.0 44674.0 40740.5 17309.0
解决方案
here is a possible solution
apply(df[,2:ncol(df)], 2, function(x){
model = lm(x ~ df$time - 1)
return(coef(model))
})
#ouput
sample1 sample2 sample3
4 3 2
It assumes the intercept is 0. If you would like the intercept evaluated use:
model = lm(x ~ df$time)
What this code does is to apply a function over columns (indicated by 2) of the data frame (df). It takes all columns from the second till the last one ([2:ncol(df]), and does a linear regression (model = lm(x ~ df$time - 1) and returns the coefficient(s) (return(coef(model)).
If the first column is not always named time then:
model = lm(x ~ df[,1] - 1)
to indicate the first column is the x
EDIT: problem is the variables were not coded as numeric. Here is a solution:
df$Time = as.numeric(df$Time) #convert time to numeric
Without intercept:
apply(df[,2:ncol(df)], 2, function(x){
x = as.numeric(x) #convert x to numeric
model = lm(x ~ df$Time - 1)
return(coef(model))
})
With intercept:
apply(df[,2:ncol(df)], 2, function(x){
x = as.numeric(x) #convert x to numeric
model = lm(x ~ df$time)
return(coef(model)[2])
})
这篇关于计算荧光随时间的斜率的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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