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Broadcasting at work: centring and scaling

Centre each station’s temperatures on its mean and scale them to standard deviation 1 with two broadcasts, fix the day means with keepdims, and build a table of distances from a column minus a row: the feature scaling and distances of k-nearest neighbours.
P02.2.4L1 Core25 min
In review

The maths and code on this page are checked by automated tests on every change, but an independent reader has not signed it off yet. Spot a mistake or a confusing step? Tell us.

Before this lesson: P02.2.3 The broadcasting rule

What you will be able to do
  • Centre the columns of a table with temps - temps.mean(axis=0), and show that every centred column sums to zero.
  • Standardise each column to mean 0 and standard deviation 1, finishing the coast column by hand.
  • Repair a failed subtraction of row means with keepdims=True or reshape(4, 1), and explain why it works.
  • Build the table of distances between points on a line as a column minus a row, and read the nearest neighbour from it.

The question

The valley is always warmer than the hill, so a reading of 23 degrees is a hot day on the hill and a cold one in the valley. How do you put every station on the same footing, so that “unusually warm” means the same thing everywhere? A k-nearest-neighbours model asks the same question of its features, and one line of broadcasting answers it.

From the last lesson

1. What shape does a (4, 3) array combined with a (3,) array give, and why?

Show answer

(4, 3). Lined up from the right, 3 matches 3; the missing axis counts as 1 and is stretched to 4.

2. What does a column (4, 1) minus a row (4,) give?

Show answer

A (4, 4) table, with no error: the column stretches across and the row stretches down.

Intuition

Subtract each station’s mean from its own column. What is left is the anomaly: how much warmer (positive) or cooler (negative) than that station’s usual the reading was. In NumPy it is the expression from the hook of P02.2.3, and the rule of that lesson says why it works: the row of station means, shape (3,), is stretched down the four days.

Three four-by-three grids with columns coast, hill, valley and rows mon, tue, wed, thu. Left, blue, temps (4, 3): 21, 19, 26; 27, 23, 28; 27, 18, 30; 29, 20, 32. A minus sign. Middle, sky blue, means (3,), stretched: a solid first row 26, 20, 29 and three dashed ghost rows repeating it. An equals sign. Right, vermillion, anomalies (4, 3): minus 5, minus 1, minus 3; 1, 3, minus 1; 1, minus 2, 1; 3, 0, 3. Under the left grid the code temps - temps.mean(axis=0); under the right grid the note each column sums to 0.
P02.2.4-F1

Station anomalies. The row of station means is stretched down the days and subtracted; each column of the result sums to 0.

Now the stations can be compared. Thursday was 3 degrees above usual at the coast and the valley, and exactly usual on the hill. Monday was below usual everywhere.

Build it up

Step 1. Centring. temps - temps.mean(axis=0) gives the anomalies in the figure. Centring makes every column average 0: the coast column \({-5 + 1 + 1 + 3}\) sums to 0, and so do the others.

Step 2. The spread of each column. The standard deviation of a column is the square root of the mean squared anomaly. temps.std(axis=0) gives [3, 1.871, 2.236] to three decimals: the coast varies most, the hill least. NumPy divides by \(n\), the number of readings; pandas divides by \(n - 1\) (P02.4.1), which gives 3.464 for the coast instead of 3.

Step 3. Standardising. Divide each anomaly by its column’s standard deviation: (temps - temps.mean(axis=0)) / temps.std(axis=0). Both operations broadcast a row of three numbers down the four days. The results are called z-scores, and every column now has mean 0 and standard deviation 1. This is the feature scaling of ML01.3.3.

Two panels of three horizontal lanes, one per station, coast, hill and valley, each with four blue dots and a vermillion tick at the mean. Left, titled Raw temperatures in degrees C, on an axis from 16 to 34: the coast dots at 21, 27 (twice, stacked) and 29 around a mean of 26; the hill dots at 18, 19, 20 and 23 around 20; the valley dots at 26, 28, 30 and 32 around 29. The note reads means 26, 20, 29. Right, titled Standardised (z), on an axis from minus 2 to 2: all three lanes have their mean tick at 0 and dots spread over about minus 1.7 to 1.6. The note reads mean 0, sd 1 for each.
P02.2.4-F2

Left: the stations sit at different levels with different spreads. Right: after standardising, each is centred at 0 with a spread of 1, so the dots can be compared directly.

Step 4. The day means need a column. Centring each day instead means subtracting temps.mean(axis=1), shape (4,), which clashes with the 3 stations. Two repairs give the means the shape (4, 1): temps.mean(axis=1, keepdims=True) keeps the collapsed axis (P02.2.2), and temps.mean(axis=1).reshape(4, 1) reshapes afterwards (P02.1.3). Lesson P02.3.1 shows a third spelling, [:, None].

Step 5. A column minus a row. Put the four day means on a line: 22, 26, 25 and 27 degrees. With m holding them, m.reshape(4, 1) - m broadcasts a (4, 1) column against a (4,) row into a (4, 4) table of every difference, and np.abs turns differences into distances. This is the “silent” broadcast of the last lesson’s pitfall, now used on purpose.

Left: the four days placed as blue dots on a temperature line from 21 to 28, mon at 22, wed at 25, tue at 26 and thu at 27, with the notes mon to wed: 3 degrees, the nearest day to mon, and the code np.abs(m.reshape(4, 1) - m), (4, 1) minus (4,) gives (4, 4). Right: a vermillion four-by-four table with rows and columns mon, tue, wed, thu: 0, 4, 3, 5; 4, 0, 1, 1; 3, 1, 0, 2; 5, 1, 2, 0. The diagonal zeros are on a pale fill and the 3 in row mon, column wed is highlighted yellow. Below: zeros on the diagonal; the same both sides.
P02.2.4-F5

The distances between the four days, computed as a column minus a row. Row mon shows that Wednesday is Monday’s nearest day, 3 degrees away.

The formal statement

NoteDefinition: centring and standardising the columns of a table

Let column \(j\) of a table hold \(x_{1j}, \dots, x_{nj}\). Its mean and its standard deviation (dividing by \(n\)) are

\[ \begin{aligned} \cparam{\mu_j} &= \frac{1}{n} \sum_{i=1}^{n} x_{ij}, \\ \cparam{\sigma_j} &= \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_{ij} - \mu_j)^2} . \end{aligned} \]

The centred value is \(x_{ij} - \mu_j\), and the standardised value, or z-score, is

\[ \cmodel{z_{ij}} = \frac{\cdata{x_{ij}} - \cparam{\mu_j}}{\cparam{\sigma_j}} . \]

In NumPy, the two numbers per column form rows of shape (3,), and broadcasting stretches them down the rows of the table: (x - x.mean(axis=0)) / x.std(axis=0).

Fact 1: a centred column sums to zero. Drop the index \(j\). Split the sum and use \({\sum_i x_i = n\mu}\):

\[ \begin{aligned} \sum_{i=1}^{n} (x_i - \mu) &= \sum_{i=1}^{n} x_i - n\mu \\ &= n\mu - n\mu = 0 . \end{aligned} \]

Fact 2: a standardised column has mean 0 and standard deviation 1. The mean of the \(z_i\) is the mean of the centred values divided by \(\sigma\), which is \(0\) by Fact 1. So the variance of the \(z_i\) is the mean of their squares:

\[ \begin{aligned} \frac{1}{n} \sum_{i=1}^{n} z_i^2 &= \frac{1}{n} \sum_{i=1}^{n} \frac{(x_i - \mu)^2}{\sigma^2} \\ &= \frac{1}{\sigma^2} \cdot \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2 \\ &= \frac{\sigma^2}{\sigma^2} = 1 . \end{aligned} \]

The last step is the definition of \(\sigma^2\). The standard deviation is the square root, also 1. (This needs \(\sigma > 0\): a column whose entries are all equal cannot be standardised.)

Fact 3: the distance table. For points \(m_1, \dots, m_k\) on a line, the table \({D_{ij} = \abs{m_i - m_j}}\) is the column m of shape (k, 1) minus the row m of shape (k,), made positive. It has zeros on its diagonal, since \({\abs{m_i - m_i} = 0}\), and it is symmetric, since \({\abs{m_i - m_j} = \abs{m_j - m_i}}\).

Worked example

Standardise the coast column by hand. The coast readings are 21, 27, 27 and 29.

Mean. \({(21 + 27 + 27 + 29) / 4 = 104 / 4 = 26}\).

Anomalies. \({21 - 26 = -5}\), then \(1\), \(1\) and \(3\). They sum to 0, as Fact 1 says.

Squares. \(25\), \(1\), \(1\) and \(9\), with sum \(36\). Their mean is \({36 / 4 = 9}\), the variance.

Standard deviation. \({\sqrt{9} = 3}\).

z-scores. Divide each anomaly by 3: \({-5/3 \approx -1.667}\), then \({1/3 \approx 0.333}\) twice, and \({3/3 = 1}\).

A trace of four columns of cells, one row per day mon, tue, wed, thu. Blue column x: 21, 27, 27, 29, with mean 26 below. Vermillion column x - 26: minus 5, 1, 1, 3, with sum 0 below. Reddish-purple column (x - 26)**2: 25, 1, 1, 9, with mean 9 below. Vermillion column (x - 26) / 3: minus 1.67, 0.33, 0.33, 1, with sd = square root of 9 = 3 below. A line at the bottom reads z: minus 1.67, 0.33, 0.33, 1 (mean 0, sd 1).
P02.2.4-F3

The coast column standardised step by step. The squares have mean 9, so the standard deviation is 3 and each anomaly is divided by 3.

Check Fact 2 on these numbers. The z-scores sum to \({-5/3 + 1/3 + 1/3 + 1 = 0}\). Their squares are \({25/9 + 1/9 + 1/9 + 9/9 = 36/9 = 4}\), and \({4 / 4 = 1}\): mean square 1, so standard deviation 1.

Code

First the station anomalies with loops, then with one broadcast.

import numpy as np

temps = np.array([[21, 19, 26],
                  [27, 23, 28],
                  [27, 18, 30],
                  [29, 20, 32]])

means = temps.mean(axis=0)
anomalies_loop = []
for i in range(4):
    row = []
    for j in range(3):
        row.append(temps[i, j] - means[j])
    anomalies_loop.append(row)

anomalies = temps - means
print(anomalies)
print(anomalies.sum(axis=0), np.array_equal(anomalies, anomalies_loop))
[[-5. -1. -3.]
 [ 1.  3. -1.]
 [ 1. -2.  1.]
 [ 3.  0.  3.]]
[0. 0. 0.] True

Now the standard deviations and the z-scores.

stds = temps.std(axis=0)
z = (temps - temps.mean(axis=0)) / temps.std(axis=0)
print(stds.round(3))
print(z.round(3))
print(z.mean(axis=0).round(10), z.std(axis=0).round(10))
print(temps[:, 0].std(ddof=1).round(3))
[3.    1.871 2.236]
[[-1.667 -0.535 -1.342]
 [ 0.333  1.604 -0.447]
 [ 0.333 -1.069  0.447]
 [ 1.     0.     1.342]]
[-0.  0.  0.] [1. 1. 1.]
3.464

The hill’s standard deviation is \({\sqrt{3.5} \approx 1.871}\) and the valley’s \({\sqrt{5} \approx 2.236}\). The column means of z print as 0 and the standard deviations as 1 after rounding; before rounding, the last digits can differ from machine to machine. ddof=1 switches NumPy to dividing by \(n - 1\), which is the 3.464 that pandas reports for the coast.

The day anomalies: the flat means fail, and either repair works.

day_means = temps.mean(axis=1)
try:
    temps - day_means
except ValueError:
    print("shapes", temps.shape, "and", day_means.shape, "do not broadcast")

by_day = temps - temps.mean(axis=1, keepdims=True)
same = temps - day_means.reshape(4, 1)
print(by_day)
print(np.array_equal(by_day, same), by_day.sum(axis=1))
shapes (4, 3) and (4,) do not broadcast
[[-1. -3.  4.]
 [ 1. -3.  2.]
 [ 2. -7.  5.]
 [ 2. -7.  5.]]
True [0. 0. 0. 0.]

Finally the distance table, and each day’s nearest other day. The diagonal is set to infinity first, so that a day is not its own nearest neighbour; argmin returns the first position on a tie.

m = temps.mean(axis=1)
dist = np.abs(m.reshape(4, 1) - m)
print(dist)
others = dist.copy()
np.fill_diagonal(others, np.inf)
print(others.argmin(axis=1))
[[0. 4. 3. 5.]
 [4. 0. 1. 1.]
 [3. 1. 0. 2.]
 [5. 1. 2. 0.]]
[2 2 1 1]

Monday’s nearest day is Wednesday (position 2). Tuesday is 1 degree from both Wednesday and Thursday, and argmin picks the first, Wednesday. k-nearest neighbours (ML01.3) does the same with many features at once: the table is built by broadcasting, and every feature is standardised first so that none dominates the distance.

Explore

The broadcast stretcher is in lesson P02.2.3. Pick (4, 3) for A and (4,) for B to see the day means fail, then (4, 1) for B to see the repair stretch across the stations.

Pitfalls

Two panels. Left, marked with a cross and titled temps - day_means: the note shapes (4, 3) and (4,), and a table lining up temps, 4 and 3, over days, a grey 1 and 4, with the result 4 and a reddish-purple cross, the notes 1 to 4 and 3 vs 4, and the text ValueError: 3 vs 4 on the right-hand axis. Right, marked with a tick and titled keepdims=True: a column: the code temps - temps.mean(axis=1, keepdims=True), a sky-blue grid with a solid first column 22, 26, 25, 27 and two dashed ghost columns labelled (4, 1), stretched, and a green grid of day anomalies minus 1, minus 3, 4; 1, minus 3, 2; 2, minus 7, 5; 2, minus 7, 5, with the note each row now sums to 0.
P02.2.4-F4

Wrong beside right. The flat day means (4,) line up with the stations and clash; kept as a column (4, 1) they stretch across the stations and every row is centred.

“The day means are just as easy to subtract as the station means.”

Only once they are a column. temps.mean(axis=1) has shape (4,), which lines up with the 3 stations and fails. Use keepdims=True or reshape(4, 1), and read the shapes before you subtract.

“np.std and pandas’ std give the same number.”

np.std divides by \(n\) by default (ddof=0) and gives 3 for the coast; pandas’ Series.std divides by \(n - 1\) and gives 3.464. Say which one you used. scikit-learn’s StandardScaler divides by \(n\), like NumPy.

“Standardise the whole dataset before splitting it.”

The means and standard deviations are themselves learned from data. Computed on all the rows, they carry information from the test rows into training, the leak of ML01.2.4. Compute them on the training rows only and apply the same two rows of numbers to every other row, as ML01.3.3 does.

“Every column can be standardised.”

A column whose entries are all equal has standard deviation 0, and dividing by it gives nan (0 divided by 0) with a warning. Check stds > 0 first, or drop the constant column: it carries no information anyway.

Check yourself

Recall

1. What are the station anomalies of Thursday?

Show answer

[3, 0, 3]: Thursday’s readings 29, 20, 32 minus the station means 26, 20, 29.

2. Why does every column of temps - temps.mean(axis=0) sum to zero?

Show answer

The sum of \(x_i - \mu\) is \({\sum x_i - n\mu}\), and \({\sum x_i = n\mu}\) by the definition of the mean.

3. What is the standard deviation of the coast column, and what are its z-scores?

Show answer

3, dividing by \(n\). The z-scores are \(-5/3\), \(1/3\), \(1/3\) and \(1\), about \(-1.667\), \(0.333\), \(0.333\), \(1\).

4. Name two ways to make temps - temps.mean(axis=1) work.

Show answer

temps - temps.mean(axis=1, keepdims=True) and temps - temps.mean(axis=1).reshape(4, 1). Both give the day means the shape (4, 1).

5. What shape does m.reshape(4, 1) - m have for m of shape (4,), and what is on its diagonal?

Show answer

(4, 4). Its diagonal holds \({m_i - m_i = 0}\).

Exercises

1. ★ Standardise the valley column, 26, 28, 30, 32, by hand.

Show solution

Mean \({116 / 4 = 29}\). Anomalies \(-3\), \(-1\), \(1\), \(3\), which sum to 0. Squares \(9\), \(1\), \(1\), \(9\), with sum \(20\) and mean \(5\), so the standard deviation is \({\sqrt{5} \approx 2.236}\). The z-scores are \({-3/\sqrt{5} \approx -1.342}\), \({-1/\sqrt{5} \approx -0.447}\), \(0.447\) and \(1.342\).

2. ★★ Write one NumPy expression that scales each column of temps to run from 0 (its smallest reading) to 1 (its largest), and compute the hill column.

Show solution

(temps - temps.min(axis=0)) / (temps.max(axis=0) - temps.min(axis=0)). Both reductions have shape (3,) and broadcast down the days. For the hill, the minimum is 18 and the maximum 23, a range of 5, so 19, 23, 18 and 20 become \({1/5 = 0.2}\), \(1\), \(0\) and \({2/5 = 0.4}\).

3. ★★★ Four points sit on a line at p = [1, 2, 4, 7]. Build the distance table np.abs(p.reshape(4, 1) - p) by hand, find each point’s nearest other point, and say how many different distances between two different points the table can hold at most for \(k\) points, and why.

Show solution

Row by row: [0, 1, 3, 6], [1, 0, 2, 5], [3, 2, 0, 3], [6, 5, 3, 0]. Ignoring the diagonal, the nearest points are: 1 to 2 (distance 1), 2 to 1 (distance 1), 4 to 2 (distance 2) and 7 to 4 (distance 3). The table has \(k^2\) entries, but the \(k\) diagonal ones are 0 and the rest come in equal pairs above and below the diagonal, so it holds at most \({k(k - 1)/2}\) different distances between different points: 6 for \(k = 4\). Here they are 1, 3, 6, 2, 5 and 3, of which 5 are different.

Summary card

Centre with one broadcast, scale with another.

The anchor figure again: temps minus the row of station means stretched down the four days gives the grid of anomalies, minus 5, minus 1, minus 3; 1, 3, minus 1; 1, minus 2, 1; 3, 0, 3.
P02.2.4-F1

Anomalies: temps minus station means.

  • Anomalies: temps - temps.mean(axis=0); each column sums to 0.
  • z-scores: (x - x.mean(axis=0)) / x.std(axis=0); mean 0, sd 1 (coast: sd 3).
  • Row means need a column: keepdims=True or reshape(4, 1).
  • Distances: np.abs(m.reshape(4, 1) - m), a (4, 4) table with zeros on the diagonal.

Sources and next steps

Sources. Each was opened and checked against this lesson on 15 November 2026.

  • NumPy developers, Broadcasting, NumPy user guide, version 2.5. The general rules used here, and the failure for shapes (4,3) and (4,).
  • NumPy developers, numpy.std and numpy.mean, NumPy reference, version 2.5. std divides by N - ddof with ddof zero by default, and ddof=1 gives the \(n - 1\) version; keepdims=True leaves the reduced axis with size one so that the result broadcasts correctly against the input.
  • NumPy developers, numpy.argmin, NumPy reference, version 2.5. Notes: with several occurrences of the minimum, the index of the first is returned.
  • scikit-learn developers, StandardScaler, scikit-learn documentation, version 1.9. Notes: a biased estimator of the standard deviation is used, equivalent to numpy.std(x, ddof=0).
  • Jake VanderPlas, Computation on Arrays: Broadcasting, Python Data Science Handbook. Section “Centering an array”: subtracting the column means X.mean(0) from a table by broadcasting, and checking that the centred means are near zero.

Next. P02.3 Indexing, slicing and views picks out rows, columns and blocks of a table, and ML01.3 k-nearest neighbours puts standardised features and distance tables to work.

This lesson unlocks

P02.3 Indexing, slicing and views ML01.3 k-nearest neighbors

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