How can a Python list be sliced such that a column is moved to being a separate element column?

Question

I have a list of the following form:

[[0, 5.1, 3.5, 1.4, 0.2],
 [0, 4.9, 3.0, 1.4, 0.2],
 [0, 4.7, 3.2, 1.3, 0.2],
 [1, 4.6, 3.1, 1.5, 0.2],
 [1, 5.0, 3.6, 1.4, 0.2],
 [1, 5.4, 3.9, 1.7, 0.4],
 [1, 4.6, 3.4, 1.4, 0.3]]

I want to slice out the first column and add it as a new element to each row of data (so at each odd position in the list), changing it to the following form:

[[5.1, 3.5, 1.4, 0.2], [0],
 [4.9, 3.0, 1.4, 0.2], [0],
 [4.7, 3.2, 1.3, 0.2], [0],
 [4.6, 3.1, 1.5, 0.2], [1],
 [5.0, 3.6, 1.4, 0.2], [1],
 [5.4, 3.9, 1.7, 0.4], [1],
 [4.6, 3.4, 1.4, 0.3], [1],]

How could I do this?

So far, I have extracted the necessary information in the following ways:

targets = [element[0] for element in dataset]
features = dataset[1:]

Not quite a duplicate, but see [here](http://stackoverflow.com/questions/34057294/flat-list-as-a-result-of-list-comprehension). — TigerhawkT3, Dec 05 '15 at 06:31

Learner · Answer 1 · 2015-12-05T08:30:29.777

Try indexing and then get flattened list- i used list comprehension for flattening.

>>>l=[[0, 5.1, 3.5, 1.4, 0.2],
 [0, 4.9, 3.0, 1.4, 0.2],
 [0, 4.7, 3.2, 1.3, 0.2],
 [1, 4.6, 3.1, 1.5, 0.2],
 [1, 5.0, 3.6, 1.4, 0.2],
 [1, 5.4, 3.9, 1.7, 0.4],
 [1, 4.6, 3.4, 1.4, 0.3]]
>>>[[i[1:],[i[0]]] for i in l]#get sliced list of lists
>>>[[[5.1, 3.5, 1.4, 0.2], [0]], [[4.9, 3.0, 1.4, 0.2], [0]], [[4.7, 3.2, 1.3, 0.2], [0]], [[4.6, 3.1, 1.5, 0.2], [1]], [[5.0, 3.6, 1.4, 0.2], [1]], [[5.4, 3.9, 1.7, 0.4], [1]], [[4.6, 3.4, 1.4, 0.3], [1]]]
>>>d=[[i[1:],[i[0]]] for i in l]
>>>[item for sublist in d for item in sublist]#flatten list d
>>>[[5.1, 3.5, 1.4, 0.2], [0], [4.9, 3.0, 1.4, 0.2], [0], [4.7, 3.2, 1.3, 0.2], [0], [4.6, 3.1, 1.5, 0.2], [1], [5.0, 3.6, 1.4, 0.2], [1], [5.4, 3.9, 1.7, 0.4], [1], [4.6, 3.4, 1.4, 0.3], [1]]

Just oneliner alternative-

[item for sublist in [[i[1:],[i[0]]] for i in l] for item in sublist] #Here l is that list

This is close, but you need to flatten the result. – DSM Dec 05 '15 at 06:21 — DSM, Dec 05 '15 at 06:21
And how about an explanation? – TigerhawkT3 Dec 05 '15 at 06:23 — TigerhawkT3, Dec 05 '15 at 06:23

score 4 · Answer 2 · answered Dec 05 '15 at 07:16

4

List comprehensions are nice but can be a bit hard to scan. Loops are still useful, especially when combined with extend:

res = []
for entry in dataset:
    res.extend([entry[1:], entry[:1]])

now:

import pprint    
pprint.pprint(res)

prints:

[[5.1, 3.5, 1.4, 0.2],
 [0],
 [4.9, 3.0, 1.4, 0.2],
 [0],
 [4.7, 3.2, 1.3, 0.2],
 [0],
 [4.6, 3.1, 1.5, 0.2],
 [1],
 [5.0, 3.6, 1.4, 0.2],
 [1],
 [5.4, 3.9, 1.7, 0.4],
 [1],
 [4.6, 3.4, 1.4, 0.3],
 [1]]

answered Dec 05 '15 at 07:16

Mike Müller

82,630
20
166
161

Best answer just because you have not used a list comprehension... `[r.extend((el[1:],el[:1])) for r in [[]] for el in dataset]; print r` – gboffi Dec 05 '15 at 08:39
@gboffi Nice try. But it does not work in Python 3 because `r` is not defined after the list comprehension. You would need two lines `res = []; [res.extend([entry[1:], entry[:1]]) for entry in dataset]`. List comprehensions with side effects are not really nice. Typically you want a useful return value. – Mike Müller Dec 05 '15 at 08:47
Well, I've learned about the side effects in P2's list comprehensions the hard way... personally I'm very happy of the new behaviour and the code in my comment was just a joke. – gboffi Dec 05 '15 at 09:05
Maybe a candidate for a code-examples-version of https://pypi.python.org/pypi/pyjokes ;). – Mike Müller Dec 05 '15 at 09:14

score 2 · Answer 3 · answered Dec 05 '15 at 06:22

Slice each sublist and make a new list with an element for each slice:

l = [[0, 5.1, 3.5, 1.4, 0.2],
 [0, 4.9, 3.0, 1.4, 0.2],
 [0, 4.7, 3.2, 1.3, 0.2],
 [1, 4.6, 3.1, 1.5, 0.2],
 [1, 5.0, 3.6, 1.4, 0.2],
 [1, 5.4, 3.9, 1.7, 0.4],
 [1, 4.6, 3.4, 1.4, 0.3]]

>>> print(*[item for sub in l for item in (sub[1:], [sub[0]])], sep='\n')
[5.1, 3.5, 1.4, 0.2]
[0]
[4.9, 3.0, 1.4, 0.2]
[0]
[4.7, 3.2, 1.3, 0.2]
[0]
[4.6, 3.1, 1.5, 0.2]
[1]
[5.0, 3.6, 1.4, 0.2]
[1]
[5.4, 3.9, 1.7, 0.4]
[1]
[4.6, 3.4, 1.4, 0.3]
[1]

Using the print on the list comprehension makes it rather difficult to scan. Using two lines and having the print separated would be easier on the eye. Also, you can use `pprint`. — Mike Müller, Dec 05 '15 at 07:20

Ahsanul Haque · Accepted Answer · 2015-12-05T07:53:39.180

2

Try this:

from itertools import chain
print list(chain(*[list((element[1:],[element[0]])) for element in a]))

Output:

[[5.1, 3.5, 1.4, 0.2], [0], [4.9, 3.0, 1.4, 0.2], [0],
 [4.7, 3.2, 1.3, 0.2], [0], [4.6, 3.1, 1.5, 0.2], [1], 
 [5.0, 3.6, 1.4, 0.2], [1], [5.4, 3.9, 1.7, 0.4], [1], 
 [4.6, 3.4, 1.4, 0.3], [1]]

edited Dec 05 '15 at 07:53

answered Dec 05 '15 at 06:33

Ahsanul Haque

10,676
4
41
57

Instead of unpacking you can use `list(itertools.chain.from_iterable([list((element[1:],[element[0]])) for element in l]))` – Learner Dec 05 '15 at 07:02

Mazdak · Answer 5 · 2015-12-05T12:46:57.050

1

A Pythonic approach in python 3.X using unpacking iteration and itertools.chain:

>>> from itertools import chain
>>> 
>>> list(chain.from_iterable([[j,[i]] for i,*j in A]))
[[5.1, 3.5, 1.4, 0.2], [0], 
 [4.9, 3.0, 1.4, 0.2], [0], 
 [4.7, 3.2, 1.3, 0.2], [0], 
 [4.6, 3.1, 1.5, 0.2], [1], 
 [5.0, 3.6, 1.4, 0.2], [1], 
 [5.4, 3.9, 1.7, 0.4], [1], 
 [4.6, 3.4, 1.4, 0.3], [1]]

edited Dec 05 '15 at 12:46

answered Dec 05 '15 at 07:54

Mazdak

105,000
18
159
188

1

This is Python3-only. The `*j` unpacking does not work in Python 2. – Mike Müller Dec 05 '15 at 09:01

How can a Python list be sliced such that a column is moved to being a separate element column?

5 Answers5