[R] differing number of rows when running nls regression

[Eric Berger]

I tried running your code. I did the following
i. I omitted the preview() statement because x_25_2024 was not provided
ii. I omitted the start=list(a=A,b=B) from the nls() statement.
iii. I ran the nls command using the form that you defined

> foo <- nls(form, data=df)

It ran with no errors.

> plot(foo)

worked.

FYI my sessionInfo()

> sessionInfo()
R version 4.5.1 (2025-06-13)
Platform: x86_64-pc-linux-gnu
Running under: Ubuntu 22.04.5 LTS

Matrix products: default
BLAS:   /usr/lib/x86_64-linux-gnu/openblas-pthread/libblas.so.3
LAPACK: /usr/lib/x86_64-linux-gnu/openblas-pthread/libopenblasp-r0.3.20.so;
 LAPACK version 3.10.0

locale:
 [1] LC_CTYPE=en_US.UTF-8       LC_NUMERIC=C
 [3] LC_TIME=en_US.UTF-8        LC_COLLATE=en_US.UTF-8
 [5] LC_MONETARY=en_US.UTF-8    LC_MESSAGES=en_US.UTF-8
 [7] LC_PAPER=en_US.UTF-8       LC_NAME=C
 [9] LC_ADDRESS=C               LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C

time zone: Asia/Jerusalem
tzcode source: system (glibc)

attached base packages:
[1] stats     graphics  grDevices datasets  utils     methods   base

other attached packages:
[1] nlstools_2.1-0 nlme_3.1-166

loaded via a namespace (and not attached):
[1] compiler_4.5.1 tools_4.5.1    bspm_0.5.7     grid_4.5.1
lattice_0.22-6

HTH,
Eric



On Sat, Sep 20, 2025 at 9:02 AM Luigi Marongiu <marongiu.luigi using gmail.com>
wrote:

> I have some data (y=Response, x=Dose) that is distributed roughly as a
> curve. I am trying to fit a non-linear regression model on the data
> using negative logistic function (even if the data show a rise in the
> final part of the curve that the logistic won't accommodate for).
> I used a rough graphical approach to guess the starting values then
> the package `nlstools` for a better overview of the data. However, I
> get the error "differing number of rows".
> If I run `nls` directly, I get the error" "Error in 1 + exp :
> non-numeric argument to binary operator"
> Why is that happening?
> How can I set the model?
> Is there a better curve function to fit the data?
> Thank you
>
> ```
> df = data.frame(Response =    c(890.72,    895.46,    895.63,
> 894.72,    895.49,    893.59,
>                 892.53,    895.06,    897.21,    889.27,    876.05,
> 857.96,    862.02,    858.36,
>                 890.94,    890.8,    887.22,    888.91,    890.83,
> 889.92,    891.76,    890.32,
>                 886.35,    878.11,    866.57,    859.04,    863.64,
> 880.16,    884.15,    879.57,
>                 878.89,    882.27,    881.59,    880.98,    881.45,
> 876.19,    868.32,    859.16,
>                 850.53,    853.21,    859.34,    859.73,    861.19),
>                 Dose =    c(0.0000000015,    0.000000003,
> 0.000000006,    0.000000012,
>                        0.000000024,    0.000000048,    0.000000095,
> 0.00000018,
>                        0.00000038,    0.00000078,    0.0000015,
> 0.000013,    0.000025,
>                        0.00005,    0.0000000015,    0.000000003,
> 0.000000006,
>                        0.000000012,    0.000000024,    0.000000048,
> 0.000000095,
>                        0.00000018,    0.00000038,    0.00000078,
> 0.0000015,    0.000025,
>                        0.00005,    0.0000000015,    0.000000003,
> 0.000000006,
>                        0.000000012,    0.000000024,    0.000000048,
> 0.000000095,
>                        0.00000018,    0.00000038,    0.00000078,
> 0.0000015,    0.000003,
>                        0.000006,    0.000013,    0.000025,    0.00005)
> )
> plot(Response~log10(Dose), df)
> abline(a=500, b=-60)
> abline(v=log10(0.3e-6))
> A = 0.3e-6     # plateau
> B = -60        # slope
> library(nlstools)   # NL regression
> form = as.formula(Response ~ ( (exp(a+b*Dose)) / (1+exp-(a+b*Dose)) ) )
> preview(form, data = x_25_2024, start=list(a=A, b=B))
> nls(Response ~ ( (exp(a+b*Dose)) / (1+exp-(a+b*Dose)) ),
>     data=df, start=list(a=A, b=B))
> ```
>
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