使用itertools groupby查找列表中所有偶数的组

Question

I am trying to understand the usefulness of the itertools.groupby() function and have created a rather naive use case. I have a list of numbers, and want to group them by oddness or evenness. Below is my code using itertools.groupby() :

for decision, group in groupby(range(2, 11), key=lambda x: x % 2 == 0): 
    ...:     print(decision, list(group))

And below is the output I get:

True [2]
False [3]
True [4]
False [5]
True [6]
False [7]
True [8]
False [9]
True [10]

Basically what I was expecting is something like where all the "True's" are grouped together and all the "False's" are grouped together.

Is this even possible with groupby() ?

Answer 1

groupby() combines consecutive values where their key output is equal, giving you a shared iterator for each such group. It will not process the whole input in one go, it gives you the groups as you iterate .

For your example, the key changes for each value, and so the groups consist of a single value each; it starts with 2 and the key output is True , then next comes 3 and the key produces False . Because True != False , that's a new group. The next value 4 changes the key again, to True , so it's another group, etc.

What you want to do can't be done with groupby() ; to sort values into buckets across the whole iterable, just use a dictionary:

grouped = {}
for value in range(2, 11):
    key = value % 2 == 0
    grouped.setdefault(key, []).append(value)

You could also sort your input first, with sorted(range(2, 11), key=lambda x: x % 2 == 0) , but that's a waste of time . Sorting is a more complex algorithm, taking O(n log n) ( quasilinear ) time, whereas using a dictionary to group the values takes O(n) ( linear ) time. That might not matter when you have just 9 elements, but when you have to process 1000 elements, sorting would increase the time taken by a factor of nearly 10, for 1 million elements, a factor of nearly 20, etc.