pr338 - 1 year ago 189

Python Question

How can I find median of a rdd of integers using a distributed method, IPython, and Spark? The rdd is approximately 700,000 elements and therefore too large to collect and find the median.

This question is similar to this question. However, the answer to the question is using Scala, which I do not know.

How can I calculate exact median with Apache Spark?

Using the thinking for the Scala answer, I am trying to write a similar answer in Python.

I know I first want to sort the rdd. I do not know how. I see the

`sortBy`

`sortByKey`

- First, I was thinking of doing ?
`myrdd.sortBy(lambda x: x)`

- Next I will find the length of the rdd ().
`rdd.count()`

- Finally, I want to find the element or 2 elements at the center of the rdd. I need help with this method too.

EDIT:

I had an idea. Maybe I can index my rdd and then key = index and value = element. And then I can try to sort by value? I don't know if this is possible because there is only a

`sortByKey`

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Answer Source

You can use `approxQuantile`

method which implements Greenwald-Khanna algorithm:

**Python**:

```
df.approxQuantile("x", [0.5], 0.25)
```

**Scala**:

```
df.stat.approxQuantile("x", Array(0.5), 0.25)
```

where the last parameter is a relative error. The lower the number the more accurate results and more expensive computation.

**Python**

As I've mentioned in the comments it is most likely not worth all the fuss. If data is relatively small like in your case then simply collect and compute median locally:

```
import numpy as np
np.random.seed(323)
rdd = sc.parallelize(np.random.randint(1000000, size=700000))
%time np.median(rdd.collect())
np.array(rdd.collect()).nbytes
```

It takes around 0.01 second on my few years old computer and around 5.5MB of memory.

If data is much larger sorting will be a limiting factor so instead of getting an exact value it is probably better to sample, collect, and compute locally. But if you really want a to use Spark something like this should do the trick (if I didn't mess up anything):

```
from numpy import floor
import time
def quantile(rdd, p, sample=None, seed=None):
"""Compute a quantile of order p ∈ [0, 1]
:rdd a numeric rdd
:p quantile(between 0 and 1)
:sample fraction of and rdd to use. If not provided we use a whole dataset
:seed random number generator seed to be used with sample
"""
assert 0 <= p <= 1
assert sample is None or 0 < sample <= 1
seed = seed if seed is not None else time.time()
rdd = rdd if sample is None else rdd.sample(False, sample, seed)
rddSortedWithIndex = (rdd.
sortBy(lambda x: x).
zipWithIndex().
map(lambda (x, i): (i, x)).
cache())
n = rddSortedWithIndex.count()
h = (n - 1) * p
rddX, rddXPlusOne = (
rddSortedWithIndex.lookup(x)[0]
for x in int(floor(h)) + np.array([0L, 1L]))
return rddX + (h - floor(h)) * (rddXPlusOne - rddX)
```

And some tests:

```
np.median(rdd.collect()), quantile(rdd, 0.5)
## (500184.5, 500184.5)
np.percentile(rdd.collect(), 25), quantile(rdd, 0.25)
## (250506.75, 250506.75)
np.percentile(rdd.collect(), 75), quantile(rdd, 0.75)
(750069.25, 750069.25)
```

Finally lets define median:

```
from functools import partial
median = partial(quantile, p=0.5)
```

So far so good but it takes 4.66 s in a local mode without any network communication. There is probably way to improve this, but why even bother?

**Language independent** (*Hive UDAF*):

If you use `HiveContext`

you can also use Hive UDAFs. With integral values:

```
rdd.map(lambda x: (float(x), )).toDF(["x"]).registerTempTable("df")
sqlContext.sql("SELECT percentile_approx(x, 0.5) FROM df")
```

With continuous values:

```
sqlContext.sql("SELECT percentile(x, 0.5) FROM df")
```

In `percentile_approx`

you can pass an additional argument which determines a number of records to use.

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