如何对日期数据执行指数拟合？
How to perform an exponential fit on date data?
我有一些这种格式的数据：
0,1/19/20
0,1/20/20
0,1/21/20
1,1/22/20
6,1/23/20
7,1/24/20
11,1/25/20
15,1/26/20
28,1/27/20
38,1/28/20
我想将它与适合的指数曲线一起绘制。
这就是我正在尝试的：
set datafile separator ","
set terminal svg enhance background rgb "white"
set timefmt "%m/%d/%y"
set xdata time
set format x "%m/%d/%y"
f(x)=a*exp(x*b)
a=1
b=0.00000000001
fit f(x) "test.dat" using 2:1 via a,b
plot "test.dat" using 2:1, f(x)
但这就是我得到的：
https://i.stack.imgur.com/JrdYq.jpg
输出：
iter chisq delta/lim lambda a b
0 2.4518484242e+03 0.00e+00 7.18e-01 1.000000e+00 1.000000e-11
1 1.0605090928e+03 -1.31e+05 7.18e-02 1.397524e+01 1.205276e-11
2 1.0548499551e+03 -5.36e+02 7.18e-03 1.473392e+01 1.761422e-11
* 1.1929533566e+03 1.16e+04 7.18e-02 5.600970e+00 4.096647e-10
3 1.0548479644e+03 -1.89e-01 7.18e-03 1.463642e+01 2.152072e-11
iter chisq delta/lim lambda a b
After 3 iterations the fit converged.
final sum of squares of residuals : 1054.85
rel. change during last iteration : -1.88724e-06
degrees of freedom (FIT_NDF) : 5
rms of residuals (FIT_STDFIT) = sqrt(WSSR/ndf) : 14.5248
variance of residuals (reduced chisquare) = WSSR/ndf : 210.97
Final set of parameters Asymptotic Standard Error
======================= ==========================
a = 14.6364 +/- 4.851e+04 (3.315e+05%)
b = 2.15207e-11 +/- 2.098e-06 (9.749e+06%)
correlation matrix of the fit parameters:
a b
a 1.000
b -1.000 1.000
另一种情况（数据格式相同，命令相同）：
剧情：https://i.stack.imgur.com/SUrOb.jpg
输出：
iter chisq delta/lim lambda a b
0 3.1449068545e+15 0.00e+00 8.39e+07 1.000000e+00 1.000000e-08
1 4.3978813150e+14 -6.15e+05 8.39e+06 9.961173e-01 9.380902e-09
2 6.0443707833e+13 -6.28e+05 8.39e+05 9.921770e-01 8.756643e-09
3 8.2818940310e+12 -6.30e+05 8.39e+04 9.880292e-01 8.132279e-09
4 1.1307879559e+12 -6.32e+05 8.39e+03 9.783880e-01 7.512240e-09
5 1.4643178088e+11 -6.72e+05 8.39e+02 7.869411e-01 7.011825e-09
...
223 4.3503512818e+08 -6.75e-01 8.39e+03 7.898096e-31 4.829813e-08
iter chisq delta/lim lambda a b
After 223 iterations the fit converged.
final sum of squares of residuals : 4.35035e+08
rel. change during last iteration : -6.75096e-06
degrees of freedom (FIT_NDF) : 55
rms of residuals (FIT_STDFIT) = sqrt(WSSR/ndf) : 2812.42
variance of residuals (reduced chisquare) = WSSR/ndf : 7.90973e+06
Final set of parameters Asymptotic Standard Error
======================= ==========================
a = 7.8981e-31 +/- 1.236e-30 (0%)
b = 4.82981e-08 +/- 2.01e-07 (416.2%)
correlation matrix of the fit parameters:
a b
a 1.000
b -1.000 1.000
line 0: warning: Too many axis ticks requested (>6e+04)
line 0: warning: Too many axis ticks requested (>6e+04)
我怀疑我没有为拟合曲线选择正确的初始参数，但我玩过很多值（尤其是 b），有时我在 real()[=37 中得到 未知类型=].
我读到在处理日期时，x 取本世纪初的秒数。如果是这种情况，那么我不确定如何使用该数字进行正确的操作。如果可能的话，我还想摆脱所有 xtics 和 y 为 0 的点。
有人可以指出我做错了什么吗？也许我没有使用正确的拟合函数？
时间数据在内部处理为自 1970 年 1 月 1 日起的秒数 00:00:00。检查 help time/date。因此，您需要在开始日期之前移动函数 f(x) 并给出合理的起始值，尤其是对于 b.
代码：
### exponential fit of time data
reset session
$Data <<EOD
0,1/19/20
0,1/20/20
0,1/21/20
1,1/22/20
6,1/23/20
7,1/24/20
11,1/25/20
15,1/26/20
28,1/27/20
38,1/28/20
EOD
set datafile separator comma
myTimeFmt = "%m/%d/%y"
set format x "%m/%d" time
# get the start date
set table $Dummy
plot $Data u (Start=timecolumn(2,myTimeFmt)) index 0 every ::0:0:0:0 w table
unset table
b=1e8 # reasonable startvalue for fitting
f(x) = a*exp((x-Start)/b*log(2))+c
set fit nolog
fit f(x) $Data u (timecolumn(2,myTimeFmt)):1 via a,b,c
End = strptime(myTimeFmt,"2/11/20")
set xrange[Start:End]
set yrange[0.5:]
set logscale y
set grid xtics, ytics
set key top left
set label 1 at graph 0.01, graph 0.85 sprintf("Doubling every: %.2f days",b/84600.)
plot $Data u (timecolumn(2,myTimeFmt)):1 w lp pt 7 lc rgb "blue" ti "Data", \
f(x) w l lc rgb "red" ti "Fit"
### end of code
结果：
我有一些这种格式的数据：
0,1/19/20
0,1/20/20
0,1/21/20
1,1/22/20
6,1/23/20
7,1/24/20
11,1/25/20
15,1/26/20
28,1/27/20
38,1/28/20
我想将它与适合的指数曲线一起绘制。 这就是我正在尝试的：
set datafile separator ","
set terminal svg enhance background rgb "white"
set timefmt "%m/%d/%y"
set xdata time
set format x "%m/%d/%y"
f(x)=a*exp(x*b)
a=1
b=0.00000000001
fit f(x) "test.dat" using 2:1 via a,b
plot "test.dat" using 2:1, f(x)
但这就是我得到的： https://i.stack.imgur.com/JrdYq.jpg
输出：
iter chisq delta/lim lambda a b
0 2.4518484242e+03 0.00e+00 7.18e-01 1.000000e+00 1.000000e-11
1 1.0605090928e+03 -1.31e+05 7.18e-02 1.397524e+01 1.205276e-11
2 1.0548499551e+03 -5.36e+02 7.18e-03 1.473392e+01 1.761422e-11
* 1.1929533566e+03 1.16e+04 7.18e-02 5.600970e+00 4.096647e-10
3 1.0548479644e+03 -1.89e-01 7.18e-03 1.463642e+01 2.152072e-11
iter chisq delta/lim lambda a b
After 3 iterations the fit converged.
final sum of squares of residuals : 1054.85
rel. change during last iteration : -1.88724e-06
degrees of freedom (FIT_NDF) : 5
rms of residuals (FIT_STDFIT) = sqrt(WSSR/ndf) : 14.5248
variance of residuals (reduced chisquare) = WSSR/ndf : 210.97
Final set of parameters Asymptotic Standard Error
======================= ==========================
a = 14.6364 +/- 4.851e+04 (3.315e+05%)
b = 2.15207e-11 +/- 2.098e-06 (9.749e+06%)
correlation matrix of the fit parameters:
a b
a 1.000
b -1.000 1.000
另一种情况（数据格式相同，命令相同）：
剧情：https://i.stack.imgur.com/SUrOb.jpg
输出：
iter chisq delta/lim lambda a b
0 3.1449068545e+15 0.00e+00 8.39e+07 1.000000e+00 1.000000e-08
1 4.3978813150e+14 -6.15e+05 8.39e+06 9.961173e-01 9.380902e-09
2 6.0443707833e+13 -6.28e+05 8.39e+05 9.921770e-01 8.756643e-09
3 8.2818940310e+12 -6.30e+05 8.39e+04 9.880292e-01 8.132279e-09
4 1.1307879559e+12 -6.32e+05 8.39e+03 9.783880e-01 7.512240e-09
5 1.4643178088e+11 -6.72e+05 8.39e+02 7.869411e-01 7.011825e-09
...
223 4.3503512818e+08 -6.75e-01 8.39e+03 7.898096e-31 4.829813e-08
iter chisq delta/lim lambda a b
After 223 iterations the fit converged.
final sum of squares of residuals : 4.35035e+08
rel. change during last iteration : -6.75096e-06
degrees of freedom (FIT_NDF) : 55
rms of residuals (FIT_STDFIT) = sqrt(WSSR/ndf) : 2812.42
variance of residuals (reduced chisquare) = WSSR/ndf : 7.90973e+06
Final set of parameters Asymptotic Standard Error
======================= ==========================
a = 7.8981e-31 +/- 1.236e-30 (0%)
b = 4.82981e-08 +/- 2.01e-07 (416.2%)
correlation matrix of the fit parameters:
a b
a 1.000
b -1.000 1.000
line 0: warning: Too many axis ticks requested (>6e+04)
line 0: warning: Too many axis ticks requested (>6e+04)
我怀疑我没有为拟合曲线选择正确的初始参数，但我玩过很多值（尤其是 b），有时我在 real()[=37 中得到 未知类型=].
我读到在处理日期时，x 取本世纪初的秒数。如果是这种情况，那么我不确定如何使用该数字进行正确的操作。如果可能的话，我还想摆脱所有 xtics 和 y 为 0 的点。
有人可以指出我做错了什么吗？也许我没有使用正确的拟合函数？
时间数据在内部处理为自 1970 年 1 月 1 日起的秒数 00:00:00。检查 help time/date。因此，您需要在开始日期之前移动函数 f(x) 并给出合理的起始值，尤其是对于 b.
代码：
### exponential fit of time data
reset session
$Data <<EOD
0,1/19/20
0,1/20/20
0,1/21/20
1,1/22/20
6,1/23/20
7,1/24/20
11,1/25/20
15,1/26/20
28,1/27/20
38,1/28/20
EOD
set datafile separator comma
myTimeFmt = "%m/%d/%y"
set format x "%m/%d" time
# get the start date
set table $Dummy
plot $Data u (Start=timecolumn(2,myTimeFmt)) index 0 every ::0:0:0:0 w table
unset table
b=1e8 # reasonable startvalue for fitting
f(x) = a*exp((x-Start)/b*log(2))+c
set fit nolog
fit f(x) $Data u (timecolumn(2,myTimeFmt)):1 via a,b,c
End = strptime(myTimeFmt,"2/11/20")
set xrange[Start:End]
set yrange[0.5:]
set logscale y
set grid xtics, ytics
set key top left
set label 1 at graph 0.01, graph 0.85 sprintf("Doubling every: %.2f days",b/84600.)
plot $Data u (timecolumn(2,myTimeFmt)):1 w lp pt 7 lc rgb "blue" ti "Data", \
f(x) w l lc rgb "red" ti "Fit"
### end of code
结果：