How do I use itertools.groupby()?

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I haven't been able to find an understandable explanation of how to actually use Python's itertools.groupby() function. What I'm trying to do is this:

  • Take a list - in this case, the children of an objectified lxml element
  • Divide it into groups based on some criteria
  • Then later iterate over each of these groups separately.

I've reviewed the documentation, and the examples, but I've had trouble trying to apply them beyond a simple list of numbers.

So, how do I use of itertools.groupby()? Is there another technique I should be using? Pointers to good "prerequisite" reading would also be appreciated.

IMPORTANT NOTE: You have to sort your data first.


The part I didn't get is that in the example construction

groups = []
uniquekeys = []
for k, g in groupby(data, keyfunc):
   groups.append(list(g))    # Store group iterator as a list
   uniquekeys.append(k)

k is the current grouping key, and g is an iterator that you can use to iterate over the group defined by that grouping key. In other words, the groupby iterator itself returns iterators.

Here's an example of that, using clearer variable names:

from itertools import groupby

things = [("animal", "bear"), ("animal", "duck"), ("plant", "cactus"), ("vehicle", "speed boat"), ("vehicle", "school bus")]

for key, group in groupby(things, lambda x: x[0]):
    for thing in group:
        print "A %s is a %s." % (thing[1], key)
    print " "

This will give you the output:

A bear is a animal. A duck is a animal.

A cactus is a plant.

A speed boat is a vehicle. A school bus is a vehicle.

In this example, things is a list of tuples where the first item in each tuple is the group the second item belongs to.

The groupby() function takes two arguments: (1) the data to group and (2) the function to group it with.

Here, lambda x: x[0] tells groupby() to use the first item in each tuple as the grouping key.

In the above for statement, groupby returns three (key, group iterator) pairs - once for each unique key. You can use the returned iterator to iterate over each individual item in that group.

Here's a slightly different example with the same data, using a list comprehension:

for key, group in groupby(things, lambda x: x[0]):
    listOfThings = " and ".join([thing[1] for thing in group])
    print key + "s:  " + listOfThings + "."

This will give you the output:

animals: bear and duck. plants: cactus. vehicles: speed boat and school bus.

itertools.groupby() in Python, Syntax: itertools.groupby(iterable, key_func) itertools.groupby() method object in python using wand library · MySQL-Connector-Python module in Python​  Syntax: itertools.groupby(iterable, key_func) Parameters: iterable: Iterable can be of any kind (list, tuple, dictionary). key: A function that calculates keys for each element present in iterable. Return type: It returns consecutive keys and groups from the iterable. If the key function is not specified or is None, key defaults to an identity

The example on the Python docs is quite straightforward:

groups = []
uniquekeys = []
for k, g in groupby(data, keyfunc):
    groups.append(list(g))      # Store group iterator as a list
    uniquekeys.append(k)

So in your case, data is a list of nodes, keyfunc is where the logic of your criteria function goes and then groupby() groups the data.

You must be careful to sort the data by the criteria before you call groupby or it won't work. groupby method actually just iterates through a list and whenever the key changes it creates a new group.

How do I use Python's itertools.groupby()?, The example that I am mentioning below from the Python docs is quite straightforward:- groups = []. uniquekeys = []. for k, g in groupby(data,  Here, lambda x: x tells groupby() to use the first item in each tuple as the grouping key. In the above for statement, groupby returns three (key, group iterator) pairs - once for each unique key. You can use the returned iterator to iterate over each individual item in that group.

Python itertools groupby, When to use groupby. It comes into picture when there is a sequence and several elements of the sequence are related. If we want to group all  How do I use Python’s itertools.groupby()? You can use groupby to group things to iterate over. You give groupby an iterable, and a optional key function/callable by which to check the items as they come out of the iterable, and it returns an iterator that gives a two-tuple of the result of the key callable and the actual items in another iterable.

A neato trick with groupby is to run length encoding in one line:

[(c,len(list(cgen))) for c,cgen in groupby(some_string)]

will give you a list of 2-tuples where the first element is the char and the 2nd is the number of repetitions.

Edit: Note that this is what separates itertools.groupby from the SQL GROUP BY semantics: itertools doesn't (and in general can't) sort the iterator in advance, so groups with the same "key" aren't merged.

How do I use Python's itertools.groupby()?, How do I use Python's itertools.groupby()? Here, lambda x: x[0] tells groupby() to use the first item in each tuple as the grouping key. Let’s use the same datastructure as defined above. groupby returns an iterator. companies_grouped_by_country is an iterator here. key argument to groupby tells the criteria using which elements of sequence should be grouped. We want to group elements of the sequence based on each country’s country key.

Another example:

for key, igroup in itertools.groupby(xrange(12), lambda x: x // 5):
    print key, list(igroup)

results in

0 [0, 1, 2, 3, 4]
1 [5, 6, 7, 8, 9]
2 [10, 11]

Note that igroup is an iterator (a sub-iterator as the documentation calls it).

This is useful for chunking a generator:

def chunker(items, chunk_size):
    '''Group items in chunks of chunk_size'''
    for _key, group in itertools.groupby(enumerate(items), lambda x: x[0] // chunk_size):
        yield (g[1] for g in group)

with open('file.txt') as fobj:
    for chunk in chunker(fobj):
        process(chunk)

Another example of groupby - when the keys are not sorted. In the following example, items in xx are grouped by values in yy. In this case, one set of zeros is output first, followed by a set of ones, followed again by a set of zeros.

xx = range(10)
yy = [0, 0, 0, 1, 1, 1, 0, 0, 0, 0]
for group in itertools.groupby(iter(xx), lambda x: yy[x]):
    print group[0], list(group[1])

Produces:

0 [0, 1, 2]
1 [3, 4, 5]
0 [6, 7, 8, 9]

How to use itertools.groupby(), k is the current grouping key, and g is an iterator that you can use to iterate over How do I use itertools. groupby()? Take a list - in this case, the children of an  I haven't been able to find an understandable explanation of how to actually use Python's itertools.groupby() function. What I'm trying to do is this: Take a list - in this case, the children of an objectified lxml element. Divide it into groups based on some criteria. Then later iterate over each of these groups separately.

Can you help me with itertools.groupby()? Entails some pandas , Can you help me with itertools.groupby()? Entails some pandas dataframes. 15 lines of code that does this already but I'm sure I can refactor it using groupby. Here, lambda x: x[0] tells groupby() to use the first item in each tuple as the grouping key. In the above for statement, groupby returns three (key, group iterator) pairs - once for each unique key. You can use the returned iterator to iterate over each individual item in that group.

How do I use Python's itertools.groupby()? - Article, What I'm trying to do is this: Take a list - in this case, the children of an objectified lxml element Divide it into groups How do I use Python's itertools.groupby()?. Groupby groups consecutive items together based on some user-specified characteristic. Each element in the resulting iterator is a tuple, where the first element ([code ]group[/code] in my example) is the "key", which is a label for that group.

groupby - itertools - Python documentation, groupby - 4 members - Make an iterator that returns consecutive keys and groups it is usually necessary to have sorted the data using the same key function). How do I use Python's itertools.groupby()? You can use groupby to group things to iterate over. You give groupby an iterable, and a optional key function/callable by which to check the items as they come out of the iterable, and it returns an iterator that gives a two-tuple of the result of the key callable and the actual items in another iterable.

Comments
  • one useful case for the would be leetcode.com/problems/string-compression
  • Is there a way to specify the groups beforehand and then not require sorting?
  • itertools usually clicks for me, but I also had a 'block' for this one. I appreciated your examples-- far clearer than docs. I think itertools tend to either click or not, and are much easier to grasp if you happen to have hit similar problems. Haven't needed this one in the wild yet.
  • @Julian python docs seem great for most stuff but when it comes to iterators, generators, and cherrypy the docs mostly mystify me. Django's docs are doubly baffling.
  • +1 for the sorting -- I didn't understand what you meant until I grouped my data.
  • @DavidCrook very late to the party but might help someone. It's probably because your array is not sorted try groupby(sorted(my_collection, key=lambda x: x[0]), lambda x: x[0])) under the assumption that my_collection = [("animal", "bear"), ("plant", "cactus"), ("animal", "duck")] and you want to group by animal or plant
  • So you read keyfunc and were like "yeah, I know exactly what that is because this documentation is quite straightforward."? Incredible!
  • I believe most people know already about this "straightforward" but useless example, since it doesn't say what kind of 'data' and 'keyfunc' to use!! But I guess you don't know either, otherwise you would help people by clarifying it and not just copy-pasting it. Or do you?
  • Technically, the docs should probably say [''.join(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D.
  • Yes. Most of the itertools docstrings are "abridged" in this way. Since all of the itertools are iterators, they must be cast to a builtin (list(), tuple()) or consumed in a loop/comprehension to display the contents. These are redundancies the author likely excluded to conserve space.