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How do I use itertools.groupby()?


Question

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, 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.

2020/07/29
1
518
7/29/2020 7:19:49 PM

Accepted Answer

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.

2020/07/29
677
7/29/2020 8:57:41 PM

itertools.groupby is a tool for grouping items.

From the docs, we glean further what it might do:

# [k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B

# [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D

groupby objects yield key-group pairs where the group is a generator.

Features

  • A. Group consecutive items together
  • B. Group all occurrences of an item, given a sorted iterable
  • C. Specify how to group items with a key function *

Comparisons

# Define a printer for comparing outputs
>>> def print_groupby(iterable, keyfunc=None):
...    for k, g in it.groupby(iterable, keyfunc):
...        print("key: '{}'--> group: {}".format(k, list(g)))
# Feature A: group consecutive occurrences
>>> print_groupby("BCAACACAADBBB")
key: 'B'--> group: ['B']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A', 'A']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A', 'A']
key: 'D'--> group: ['D']
key: 'B'--> group: ['B', 'B', 'B']

# Feature B: group all occurrences
>>> print_groupby(sorted("BCAACACAADBBB"))
key: 'A'--> group: ['A', 'A', 'A', 'A', 'A']
key: 'B'--> group: ['B', 'B', 'B', 'B']
key: 'C'--> group: ['C', 'C', 'C']
key: 'D'--> group: ['D']

# Feature C: group by a key function
>>> # islower = lambda s: s.islower()                      # equivalent
>>> def islower(s):
...     """Return a True if a string is lowercase, else False."""   
...     return s.islower()
>>> print_groupby(sorted("bCAaCacAADBbB"), keyfunc=islower)
key: 'False'--> group: ['A', 'A', 'A', 'B', 'B', 'C', 'C', 'D']
key: 'True'--> group: ['a', 'a', 'b', 'b', 'c']

Uses

Note: Several of the latter examples derive from Víctor Terrón's PyCon (talk) (Spanish), "Kung Fu at Dawn with Itertools". See also the groupby source code written in C.

* A function where all items are passed through and compared, influencing the result. Other objects with key functions include sorted(), max() and min().


Response

# OP: Yes, you can use `groupby`, e.g. 
[do_something(list(g)) for _, g in groupby(lxml_elements, criteria_func)]
2020/07/11

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.

2020/01/24

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.

2017/08/08

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]
2013/01/21

WARNING:

The syntax list(groupby(...)) won't work the way that you intend. It seems to destroy the internal iterator objects, so using

for x in list(groupby(range(10))):
    print(list(x[1]))

will produce:

[]
[]
[]
[]
[]
[]
[]
[]
[]
[9]

Instead, of list(groupby(...)), try [(k, list(g)) for k,g in groupby(...)], or if you use that syntax often,

def groupbylist(*args, **kwargs):
    return [(k, list(g)) for k, g in groupby(*args, **kwargs)]

and get access to the groupby functionality while avoiding those pesky (for small data) iterators all together.

2014/12/28

Source: https://stackoverflow.com/questions/773
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