# [Numpy-discussion] very simple iteration question.

Wed Apr 30 03:40:44 CDT 2008

```Hi Alex,

a g wrote:
> Hi.  This is a very basic question, sorry if it's irritating.  If i
> point me to it.  That'd be great.

You should look at any of the documents below and read up on array
slicing. It is perhaps the most important and pervasive concept of Numpy
and should be understood by all users.

Numpy Tutorial: http://www.scipy.org/Tentative_NumPy_Tutorial
Numpy for MATLAB users: http://www.scipy.org/NumPy_for_Matlab_Users
Guide to Numpy

> OK: how do i iterate over an axis other than 0?
>
> I have a 3D array of data[year, week, location].  I want to iterate
> over each year at each location and run a series of stats on the
> columns (on the 52 weeks in a particular year at a particular location).
>  'for years in data:' will get the first one, but then how do i not
> iterate over the 1 axis and iterate over the 2 axis instead?

It is not clear to me whether you want to slice or iterate over an
array. Assuming you are fixing the year and location, the following code
iterates over data for fixed year and location.

for week in xrange(0, 52):
<do something with> data[year, week, loc]

Slicing is more efficient and you should use it if you can. Fixing the
year and location, the following computes the mean and standard
deviation across all weeks. All of the statements below yield scalars.

data[year, :, loc].mean() -- takes the mean of the data across weeks
data[year, :, loc].std() -- takes the standard deviation of the
data across weeks

You should download IPython and type help(numpy.array) to see one set of
functions you can call on the result of a slice (sum, min, etc.).

Although I don't know what statistics you are computing for sure, the
following code might be useful since it computes a statistic across all
weeks for each year and location value.

data.mean(axis=1)

It yields a num_years by num_locations array mu where mu[y, l] is the
average data value across all weeks for year y and loc l.

I hope this helps.

Damian
```