# How to look at only the 3rd value in all lists in a list

## Question:

I have a list of lists and I want to be able to refer to the 1st, 2nd, 3rd, etc. column in a list of lists. Here is my code for the list:

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

I want to be able to say something like:

```
matrix = [
[0, 0, 0, 5, 0, 0, 0, 0, 6],
[8, 0, 0, 0, 4, 7, 5, 0, 3],
[0, 5, 0, 0, 0, 3, 0, 0, 0],
[0, 7, 0, 8, 0, 0, 0, 0, 9],
[0, 0, 0, 0, 1, 0, 0, 0, 0],
[9, 0, 0, 0, 0, 4, 0, 2, 0],
[0, 0, 0, 9, 0, 0, 0, 1, 0],
[7, 0, 8, 3, 2, 0, 0, 0, 5],
[3, 0, 0, 0, 0, 8, 0, 0, 0],
]
if (The fourth column in this matrix does not have any 1's in it):
(then do something)
```

I want to know what the python syntax would be for the stuff in parenthesis.

## Answers:

The standard way to perform what you asked is to do a list comprehension

if (The fourth column in this matrix does not have any 1’s in it):

translates in:

```
>>>if not any([1 == row[3] for row in matrix])
```

However, depending on how often you need to perform this operation, how big is your matrix, etc… you might wish to look into numpy as it is easier (and remarkably faster) to address columns. An example:

```
>>> import numpy as np
>>> matrix = np.random.randint(0, 10, (5, 5))
>>> matrix
array([[3, 0, 9, 9, 3],
[5, 7, 7, 7, 6],
[5, 4, 6, 2, 2],
[1, 3, 5, 0, 5],
[3, 9, 7, 8, 6]])
>>> matrix[..., 3] #fourth column
array([9, 7, 2, 0, 8])
```

```
if 1 in [row[3] for row in matrix]:
```

Try this:

```
if all(row[3] != 1 for row in matrix):
# do something
```

The `row[3]`

part takes a look at the fourth element of a row, the `for row in matrix`

part looks at all the rows in the matrix – this produces a list with all the fourth elements in all the rows, that is, the whole fourth column. Now if it is true for all the elements in the fourth column that they’re different from one, then the condition is satisfied and you can do what you need inside the `if`

.

A more traditional approach would be:

```
found_one = False
for i in xrange(len(matrix)):
if matrix[i][3] == 1:
found_one = True
break
if found_one:
# do something
```

Here I’m iterating over all the rows (`i`

index) of the fourth column (`3`

index), and checking if an element is equal to one: `if matrix[i][3] == 1:`

. Notice that the `for`

cycle goes from the `0`

index up to the “height” of the matrix minus one, that’s what the `xrange(len(matrix))`

part says.