In [14]:

```
import time
print('Last updated: %s' %time.strftime('%d/%m/%Y'))
```

This IPython notebook can be found in the GitHub repository rasbt/algorithms_in_ipython_notebooks

^{ 1} is a convenient way to quickly identify outliers in datasets that only contains a small number of observations: typically $3 \leq n \leq 10$.

If we want to use this test to legitimately remove (potential) outliers from a dataset, we should keep in mind that

our data has to be normal distributed,

and that we are not supposed to use this test more than once the same data set.

In my opinion, the Dixon Q-test should only be used with great caution, since this simple statistic is based on the assumption that the data is normal distributed, which can be quite challenging to predict for small sample sizes (if no prior/additional information is provided).

Personally, I would use the Dixon Q-test to only **detect** outliers and not to **remove** those, which can help with the identification of uncertainties in the data set or problems in experimental procedures.
Intuitively, this is quite similar to an approach of identifying samples that have a large standard deviation.

For example, if I tested ~1000 chemical compounds in some sort of activity assay - each compound 5 times, I would mark compounds that contain Q-test outliers for re-testing, because there might have been some problem in the measurement procedure that could have caused this inconsistency.

First, we arange the data for our sample in ascending order (from the lowest to the highest value):

$x_1 < x_2 < . . . < x_N$

Next, we calculate the experimental Q-value ($Q_{exp}$).

Note that in a later paper in 1953, Dixon and Dean^{ 3} revisted the calculation of the Q-value and reported different equations for different scenarios:

(from: Rorabacher, David B., 1991^{ 2})

- $r_{10} \;\; for \;\; 3 \geq n \leq 7$
- $r_{11} \;\; for \;\; 8 \geq n \leq 10$
- $r_{21} \;\; for \;\; 11 \geq n \leq 13$
- $r_{22} \;\; for \;\; n \leq 14$

However, according to a statement/observation in a more recent paper (Rorabacher, David B., 1991^{ 2}): "The *r _{l0}* ratio is commonly designated as 'Q' and is generally considered to be the most convenient, legitimate, statistical test available for the rejection of deviant values from a small sample conforming to a Gaussian distribution. (It is equally well suited to larger data sets if only one outlier is present.)"

Therefore, I will use *r _{l0}* for the following implementation of the Dixon Q-test:

\begin{equation} Q_{exp} = \frac{x_{2} - x_{1}}{x_{n} - x_{1}} \end{equation}

where it is assumed that the data is arranged in ascending order: $x_1 < x_2 < . . . < x_N$

^{2} Rorabacher, David B. (1991) “Statistical Treatment for Rejection of Deviant Values: Critical Values of Dixon’s‘ Q’ Parameter and Related Subrange Ratios at the 95% Confidence Level.” Analytical Chemistry 63, no. 2 (1991): 139–46.

^{3} W.J. Dixon: "Processing data for outliers Reference": J. Biometrics 9 (1953) 74-89

If the calculated Q-value for a particular observation is larger than the critical Q-value ($Q_{exp}$ > $Q_{crit}$), this observation is considered to be an outlier according to the Q-test.

**r _{10} Critical values for Dixon's two-tailored Q-Test for 3 different confidence levels**

N | Q90% | Q95% | Q99% |
---|---|---|---|

3 | 0.941 | 0.97 | 0.994 |

4 | 0.765 | 0.829 | 0.926 |

5 | 0.642 | 0.71 | 0.821 |

6 | 0.56 | 0.625 | 0.74 |

7 | 0.507 | 0.568 | 0.68 |

8 | 0.468 | 0.526 | 0.634 |

9 | 0.437 | 0.493 | 0.598 |

10 | 0.412 | 0.466 | 0.568 |

11 | 0.392 | 0.444 | 0.542 |

12 | 0.376 | 0.426 | 0.522 |

13 | 0.361 | 0.41 | 0.503 |

14 | 0.349 | 0.396 | 0.488 |

15 | 0.338 | 0.384 | 0.475 |

16 | 0.329 | 0.374 | 0.463 |

17 | 0.32 | 0.365 | 0.452 |

18 | 0.313 | 0.356 | 0.442 |

19 | 0.306 | 0.349 | 0.433 |

20 | 0.3 | 0.342 | 0.425 |

21 | 0.295 | 0.337 | 0.418 |

22 | 0.29 | 0.331 | 0.411 |

23 | 0.285 | 0.326 | 0.404 |

24 | 0.281 | 0.321 | 0.399 |

25 | 0.277 | 0.317 | 0.393 |

26 | 0.273 | 0.312 | 0.388 |

27 | 0.269 | 0.308 | 0.384 |

28 | 0.266 | 0.305 | 0.38 |

29 | 0.263 | 0.301 | 0.376 |

30 | 0.26 | 0.29 | 0.372 |

Let's consider the following sample consisting of 5 observations:

0.142, 0.153, 0.135, 0.002, 0.175

First, we sort it in ascending order: 0.002, 0.135, 0.142, 0.153, 0.175

Next, we calculate the Q-value:

\begin{equation} Q = \frac{0.135 - 0.002}{0.175-0.002} \approx 0.7687 \end{equation}

- Now, we look up the critical value for n=5 for a confidence level 95% in the Q-table $\Rightarrow 0.71$

and we conclude that 0.002 (since 0.7687 > 0.71), that the observation 0.002 is an outlier at a confidence level of 95% according to Dixon's Q-test.

We will build a simple set of dictionaries for different confidence intervals from the tabulated data in David B. Rorabacher's paper:

Rorabacher, David B. (1991) “Statistical Treatment for Rejection of Deviant Values: Critical Values of Dixon’s‘ Q’ Parameter and Related Subrange Ratios at the 95% Confidence Level.” Analytical Chemistry 63, no. 2 (1991): 139–46.

which we will use to look up the critical Q-values (dictionary values) for different sample sizes (dictionary keys).

In [1]:

```
q90 = [0.941, 0.765, 0.642, 0.56, 0.507, 0.468, 0.437,
0.412, 0.392, 0.376, 0.361, 0.349, 0.338, 0.329,
0.32, 0.313, 0.306, 0.3, 0.295, 0.29, 0.285, 0.281,
0.277, 0.273, 0.269, 0.266, 0.263, 0.26
]
q95 = [0.97, 0.829, 0.71, 0.625, 0.568, 0.526, 0.493, 0.466,
0.444, 0.426, 0.41, 0.396, 0.384, 0.374, 0.365, 0.356,
0.349, 0.342, 0.337, 0.331, 0.326, 0.321, 0.317, 0.312,
0.308, 0.305, 0.301, 0.29
]
q99 = [0.994, 0.926, 0.821, 0.74, 0.68, 0.634, 0.598, 0.568,
0.542, 0.522, 0.503, 0.488, 0.475, 0.463, 0.452, 0.442,
0.433, 0.425, 0.418, 0.411, 0.404, 0.399, 0.393, 0.388,
0.384, 0.38, 0.376, 0.372
]
Q90 = {n:q for n,q in zip(range(3,len(q90)+1), q90)}
Q95 = {n:q for n,q in zip(range(3,len(q95)+1), q95)}
Q99 = {n:q for n,q in zip(range(3,len(q99)+1), q99)}
```

Below, I wrote some simple Python code to test one data row for Dixon Q-test outliers:

In [2]:

```
def dixon_test(data, left=True, right=True, q_dict=Q95):
"""
Keyword arguments:
data = A ordered or unordered list of data points (int or float).
left = Q-test of minimum value in the ordered list if True.
right = Q-test of maximum value in the ordered list if True.
q_dict = A dictionary of Q-values for a given confidence level,
where the dict. keys are sample sizes N, and the associated values
are the corresponding critical Q values. E.g.,
{3: 0.97, 4: 0.829, 5: 0.71, 6: 0.625, ...}
Returns a list of 2 values for the outliers, or None.
E.g.,
for [1,1,1] -> [None, None]
for [5,1,1] -> [None, 5]
for [5,1,5] -> [1, None]
"""
assert(left or right), 'At least one of the variables, `left` or `right`, must be True.'
assert(len(data) >= 3), 'At least 3 data points are required'
assert(len(data) <= max(q_dict.keys())), 'Sample size too large'
sdata = sorted(data)
Q_mindiff, Q_maxdiff = (0,0), (0,0)
if left:
Q_min = (sdata[1] - sdata[0])
try:
Q_min /= (sdata[-1] - sdata[0])
except ZeroDivisionError:
pass
Q_mindiff = (Q_min - q_dict[len(data)], sdata[0])
if right:
Q_max = abs((sdata[-2] - sdata[-1]))
try:
Q_max /= abs((sdata[0] - sdata[-1]))
except ZeroDivisionError:
pass
Q_maxdiff = (Q_max - q_dict[len(data)], sdata[-1])
if not Q_mindiff[0] > 0 and not Q_maxdiff[0] > 0:
outliers = [None, None]
elif Q_mindiff[0] == Q_maxdiff[0]:
outliers = [Q_mindiff[1], Q_maxdiff[1]]
elif Q_mindiff[0] > Q_maxdiff[0]:
outliers = [Q_mindiff[1], None]
else:
outliers = [None, Q_maxdiff[1]]
return outliers
```

Some simple assertion tests to make sure that the Dixon Q-test function behaves as expected/desired.

In [3]:

```
test_data1 = [0.142, 0.153, 0.135, 0.002, 0.175]
test_data2 = [0.542, 0.153, 0.135, 0.002, 0.175]
assert(dixon_test(test_data1) == [0.002, None]), 'expect [0.002, None]'
assert(dixon_test(test_data1, right=False) == [0.002, None]), 'expect [0.002, None]'
assert(dixon_test(test_data2) == [None, None]), 'expect [None, None]'
assert(dixon_test(test_data2, q_dict=Q90) == [None, 0.542]), 'expect [None, 0.542]'
print('ok')
```

In "real" application I would prefer `NumPy`

and/or `pandas`

, however for this simple case the in-built Python `csv`

library should suffice.

Below the example CSV file is shown that we are going to read in:

In [4]:

```
%%writefile ../../data/dixon_test_in.csv
,x1,x2,x3,x4,x5
id1,0.95,-0.65,0.6,0.82,NaN
id2,2.08,NaN,-1.43,0.38,NaN
id3,-0.46,NaN,-1.25,-2.62,0.22
id4,0.24,1.88,-0.49,-0.73,-0.49
id5,-1.65,2.1,-0.09,NaN,0.8
id6,-0.44,0.93,0.19,-4.36,-0.88
id7,0.36,-0.47,NaN,0.4,2.12
id8,1.29,-0.48,-0.6,-0.38,0.27
id9,-1.25,-1.35,1.13,1.7,-0.81
id10,0.04,1.98,NaN,NaN,NaN
```

In [5]:

```
import csv
def csv_to_list(csv_file, delimiter=','):
"""
Reads in a CSV file and returns the contents as list,
where every row is stored as a sublist, and each element
in the sublist represents 1 cell in the table.
"""
with open(csv_file, 'r') as csv_con:
reader = csv.reader(csv_con, delimiter=delimiter)
return list(reader)
def print_csv(csv_content):
""" Prints CSV file to standard output."""
print(50*'-')
for row in csv_content:
row = [str(e) for e in row]
print('\t'.join(row))
print(50*'-')
def convert_cells_to_floats(csv_cont):
"""
Converts cells to floats if possible
(modifies input CSV content list).
"""
for row in range(len(csv_cont)):
for cell in range(len(csv_cont[row])):
try:
csv_cont[row][cell] = float(csv_cont[row][cell])
except ValueError:
pass
```

In [6]:

```
csv_cont = csv_to_list('../../data/dixon_test_in.csv')
convert_cells_to_floats(csv_cont)
print_csv(csv_cont)
```

Now, let us add a new `outlier`

column and apply the Dixon Q-test function to our data set.

In [7]:

```
import math
csv_cont[0].append('outlier')
for row in csv_cont[1:]: # skips header
nan_removed = [i for i in row[1:] if not math.isnan(i)]
if len(nan_removed) >= 3:
row.append(dixon_test(nan_removed, left=True, right=True, q_dict=Q90))
else:
row.append('NaN')
```

In [8]:

```
print_csv(csv_cont)
```

As we can see in the table above, we have 2 potential outliers in our data set.

Finally, we let us write the results to a new CSV file for future reference:

In [9]:

```
def write_csv(dest, csv_cont):
""" Writes a comma-delimited CSV file. """
with open(dest, 'w') as out_file:
writer = csv.writer(out_file, delimiter=',')
for row in csv_cont:
writer.writerow(row)
write_csv('../../data/dixon_test_out.csv', csv_cont)
```

To get a visual impression of how our data looks like, let us make some simple plots.

In [10]:

```
%matplotlib inline
```

In [12]:

```
import numpy as np
from matplotlib import pyplot as plt
all_means = [np.nanmean(row[1:6]) for row in csv_cont[1:]]
all_stddevs = [np.nanstd(row[1:6]) for row in csv_cont[1:]]
fig = plt.figure(figsize=(8,6))
y_pos = np.arange(len(csv_cont[1:]))
y_pos = [x for x in y_pos]
plt.yticks(y_pos, [row[0] for row in csv_cont[1:]], fontsize=10)
plt.xlabel('measurement x')
t = plt.title('Bar plot with standard deviation')
plt.grid()
plt.barh(y_pos, all_means, xerr=all_stddevs, align='center', alpha=0.4, color='g')
plt.show()
```

A more useful plot in my opinion is Tukey's boxplot^{ 4}. Boxplots are in facts one of my preferred approaches to quickly and visually indicate outliers in a Gaussian data set. However, also boxplots have to be used with real caution and might also not very informative for small sample sizes.

Q1 = Quartile 1

^{ 4} Robert McGill, John W. Tukey and Wayne A. Larsen: "The American Statistician"
Vol. 32, No. 1 (Feb., 1978), pp. 12-16

In [13]:

```
csv_nonan = [[x for x in row[1:6] if not math.isnan(x)] for row in csv_cont[1:]]
fig = plt.figure(figsize=(8,6))
plt.boxplot(csv_nonan,0,'rs',0)
plt.yticks([y+1 for y in y_pos], [row[0] for row in csv_cont[1:]])
plt.xlabel('measurement x')
t = plt.title('Box plot')
plt.show()
```

**I really don't want to draw any conclusion about which approach is right or wrong here, since in my opinion, drawing any conclusion from a data set that is based on such a small number of observations simply just doesn't make sense!**

So you may wonder why I wasted your time if you read this article up to this point? Since Dixon's Q-test is still quite popular in certain scientific fields (e.g., chemistry) that it is important to understand its principles in order to draw your own conclusion of the presented research data that you might stumble upon in research articles or scientific talks.