I have a (pretty noisy) 1D array of data in Python that I fit a linear function to -- see plot below. The x-axis is a wavelength scale. enter image description here

I would like to measure the "goodness of fit" of this new line to my actual data at multiple different intervals in my data, because I'm trying to see if it's a worse fit in a particular region over another.

I started off by just computing the chi squared value with SciPy in different regions:

chisq = scipy.stats.chisquare(data, fit_line) 

But I got negative values, which doesn't make sense in terms of a chi squared value... however this arises because my data (and hence best fit line) is all negative. I then came across the answer here regarding the R^2 approach, but I do not know how to interpret this.

Also, the different regions I want to calculate this for have different length arrays, so I'm wondering if my results will be skewed because of this. Can I just calculate (data-fit_line)**2 for each of my data points in the region, then maybe divide by the number of points in that region to normalize it? I do not understand why the formal chisquare function by scipy is different. What would the best measurement be for my case?

  • $\begingroup$ Why not just do a linear regression? The Statsmodels library comes with a whole bunch of statistics about the fit. $\endgroup$ Commented May 9, 2019 at 20:15

2 Answers 2


Compute and plot a local goodness of fit measure.

A quick and easy method, that should apply to many such settings, is to examine a local average absolute deviation between the data and their fit. An example appears in the top row of the next figure: the data are on the left and their residuals $r_i$ (deviations) are plotted on the right.

![enter image description here

To obtain a local indication of how close the fit is, I recommend smoothing a transformed absolute residual plot of $|r_i|^{2/3}$ (loosely termed a "root residual") using a robust technique such as Loess, as was done here. Some judgement concerning how aggressively to smooth the residuals is needed. This ought to be informed by your objectives and knowledge of the data. I asked the smoother to focus on windows of data about one-quarter the total width. The root residuals and this smooth are shown at the bottom left. The colors indicate the signs of the original residuals (blue for negative, tan for positive). The smooth is the black curve.

The places where the smooth is relatively high, such as around 5600 and 6200, indicate wavelengths where the fit is not so good. (However, I wouldn't pay much attention to any peaks that are lower than about twice the average. The residuals in this example were randomly generated, and so the peaks and valleys you see here illustrate what you ought to expect for a good fit generally.)

The reason for using the two-thirds power of the absolute residuals is to help suppress the effects of individual outlying data (such as the low response near 6200). Another reason is that the squared residuals $r_i^2$ often will approximately have a chi-squared distribution and the cube root (corresponding to $(r_i^2)^{1/3} = |r_i|^{2/3}$) will tend to make it symmetrical, thereby favoring neither high nor low excursions around the average magnitude. (This is the Wilson-Hilferty transformation.)

Finally, you can simply accumulate a running total of squared residuals, as shown in the bottom right corner. Places where this plot is flatter indicate regions where the original fit is better. The utility of this plot in this example is lessened by the effect of that outlying value around 6200: the vertical axis must expand to accommodate its effect, suppressing the visual differences in slopes throughout the rest of the plot. That's why the smoothed representation of the slopes of this plot at the bottom left may be preferable.

The following is the R code used to generate the illustrations. You could use it as the start of a procedure to analyze any data set.

# Generate random data.
par(mfrow = c(2,2))

X <- data.frame(lambda = seq(4500, 7200, by = 50))
delta <- rnorm(nrow(X), 0, 6e-4)
X$amplitude <- -0.0075 + 4e-3 / 2700 * (X$lambda - 5900) + delta

plot(X, type = "s", col = "lightblue", lwd = 2,
     xlab = "Wavelength", ylab = "Value")
# Fit a model and overplot it.
fit <- lm(amplitude ~ lambda, X)
abline(fit, col = "tan", lwd = 2)
# Plot the residuals.
X$residual <- residuals(fit)
plot(X$lambda, X$residual, type = "s", col = "lightblue", lwd = 2, 
 xlab = "Wavelength", ylab = "Residual")
abline (h = 0, col = "gray", lwd = 2)
# Plot the re-expressed absolute residuals as bars.
plot(X$lambda, (X$residual^2)^(1/3), type = "h", xlab = "Wavelength", 
 ylab = "Root residual",
 col = ifelse(X$residual > 0, "tan", "lightblue"), lwd = 5, lend = 1)
# Smooth this plot.
p <- 2/3 # The power of the absolute residual to use
s <- loess(abs(residual)^p ~ lambda, X, span = 1/4)
Y <- data.frame(lambda = seq(min(X$lambda), max(X$lambda), length.out = 501))
Y$y.hat <- predict(s, Y)
with(Y, lines(lambda, y.hat, lwd = 2))
# Alternatively, plot the cumulative sum of squared residuals.
X$z <- cumsum(X$residual^2)
plot(X$lambda, X$z, type = "b", xlab = "Wavelength", ylab = "Cumulative squared residual")

par(mfrow = c(1,1))

So first off, that isn't how the the chi-square function should be used.

If you want to know the "goodness of fit", use the R squared stat. R squared tells you how much of the observed variance in the outcome is explained by the input. Here is an example in python.

import numpy as np
from sklearn.linear_model import LinearRegression
import matplotlib.pyplot as plt
from sklearn.metrics import r2_score
#Your data
N = 1000
x = np.random.normal(size = N)
y = 2*x+1 + np.random.normal(size = N)

#Your model

#Your statsitic
r2_score(y, model.predict(x.reshape(-1,1)))

This returns 0.801, so 80.1% percent of the variance in y seems to be explained by x. The higher this number the better.

As for assessing fit over different regions, I think you can look at the residuals of your model and see if there are patterns or heterogeneity of variance.

  • 2
    $\begingroup$ I don't think $R^2$ will answer this question, for two reasons: first, it doesn't measure goodness of fit in the sense described by the question; and second, it won't provide the local information that is requested. $\endgroup$
    – whuber
    Commented May 9, 2019 at 20:35

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