如何在Python中索引嵌套列表？

Question

I have a nested list as shown below:

A = [('a', 'b', 'c'),
     ('d', 'e', 'f'),
     ('g', 'h', 'i')]

and I am trying to print the first element of each list using the code:

A = [('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i')]
print A[:][0]

But I get the following output:

('a', 'b', 'c')

Required output:

('a', 'd', 'g')

How to get this output in Python?

Answer 1

A[:] just creates a copy of the whole list, after which you get element 0 of that copy.

You need to use a list comprehension here:

[tup[0] for tup in A]

to get a list, or use tuple() with a generator expression to get a tuple:

tuple(tup[0] for tup in A)

Demo:

>>> A = [('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i')]
>>> [tup[0] for tup in A]
['a', 'd', 'g']
>>> tuple(tup[0] for tup in A)
('a', 'd', 'g')

Answer 2

You can transpose a list of lists/tuples with zip(*list_of_lists) then select the items you want.

>>> a
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i')]
>>> b = zip(*a)
>>> b
[('a', 'd', 'g'), ('b', 'e', 'h'), ('c', 'f', 'i')]
>>> b[0]
('a', 'd', 'g')
>>> 
>>> c = zip(*a)[0]
>>> c
('a', 'd', 'g')
>>>

Answer 3

Python lists don't work very well as multi-dimensional arrays. If you're willing to add an extra dependency(eg if you're going to do a lot of array manipulation), numpy allows you to use the almost the exact syntax you're looking for

import numpy as np
A = np.array([('a', 'b', 'c'),
              ('d', 'e', 'f'),
              ('g', 'h', 'i')])

---

This outputs the row as an np.array(which can be accessed like a list):

>>> A[:,0]
array(['a', 'd', 'g'])

---

To get the first row as a tuple:

>>> tuple(A[:,0])
('a', 'd', 'g')

Answer 4

You can also do it this way:

>>> A = [('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i')]
>>> map(lambda t:t[0], A)
['a', 'd', 'g']
>>> tuple(map(lambda t:t[0],A))
('a', 'd', 'g')

Answer 5

You can also get the behavior you want using pandas as follows:

In [1]: import pandas as pd

In [2]: A = [('a', 'b', 'c'),
     ('d', 'e', 'f'),
     ('g', 'h', 'i')]

In [3]: df = pd.DataFrame(A)

In [4]: df[:][0]
Out[4]:
0    a
1    d
2    g
Name: 0, dtype: object

In [5]: df[:][0].values
Out[5]: array(['a', 'd', 'g'], dtype=object)