When I check the shape of an array using numpy.shape(), I sometimes get (length,1) and sometimes (length,). It looks like the difference is a column vs. row vector... but It doesn't seem like that changes anything about the array itself [except some functions complain when I pass an array with shape (length,1)].
What is the difference between these two?
Why isn't the shape just, (length)?
header('content-type:text/plain');
$arr = array(
'a' => array(
'b' => array(
'c' => 'hello',
),
),
'd' => array(
'e' => array(
'f' => 'world',
),
),
);
//prime the stack using our format
$stack = array();
foreach ($arr as $k => &$v) {
$stack[] = array(
'keyPath' => array($k),
'node' => &$v
);
}
$lookup = array();
while ($stack) {
$frame = array_pop($stack);
$lookup[join('/', $frame['keyPath'])] = &$frame['node'];
if (is_array($frame['node'])) {
foreach ($frame['node'] as $key => &$node) {
$keyPath = array_merge($frame['keyPath'], array($key));
$stack[] = array(
'keyPath' => $keyPath,
'node' => &$node
);
$lookup[join('/', $keyPath)] = &$node;
}
}
}
var_dump($lookup);
// check functionality
$lookup['a'] = 0;
$lookup['d/e/f'] = 1;
var_dump($arr);
Alternatively you could have done stuff like this to get a reference /w array_pop functionality
end($stack);
$k = key($stack);
$v = &$stack[$k];
unset($stack[$k]);
That works because php arrays have elements ordered by key creation time. unset() deletes the key, and thus resets the creation time for that key.
Problem is you're using numpy.math.cos here, which expects you to pass a scalar. Use numpy.cos if you want to apply cos to an iterable.
In [30]: import numpy as np
In [31]: np.cos(np.array([1, 2, 3]))
Out[31]: array([ 0.54030231, -0.41614684, -0.9899925 ])
Error:
In [32]: np.math.cos(np.array([1, 2, 3]))
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-32-8ce0f3c0df04> in <module>()
----> 1 np.math.cos(np.array([1, 2, 3]))
TypeError: only length-1 arrays can be converted to Python scalars
In [42]: some_df = pd.DataFrame(columns=['A'])
...: for i in range(4):
...: some_df.loc[i] = [np.random.randint(0,10,(1,3))]
...:
In [43]: some_df
Out[43]:
A
0 [[7, 0, 9]]
1 [[3, 6, 8]]
2 [[9, 7, 6]]
3 [[1, 6, 3]]
The numpy values of the column are an object dtype array, containing arrays:
In [44]: some_df['A'].to_numpy()
Out[44]:
array([array([[7, 0, 9]]), array([[3, 6, 8]]), array([[9, 7, 6]]),
array([[1, 6, 3]])], dtype=object)
If those arrays all have the same shape, stack does a nice job of concatenating them on a new dimension:
In [45]: np.stack(some_df['A'].to_numpy())
Out[45]:
array([[[7, 0, 9]],
[[3, 6, 8]],
[[9, 7, 6]],
[[1, 6, 3]]])
In [46]: _.shape
Out[46]: (4, 1, 3)
This only works with one column. stack like all concatenate treats the input argument as an iterable, effectively a list of arrays.
In [48]: some_df['A'].to_list()
Out[48]:
[array([[7, 0, 9]]),
array([[3, 6, 8]]),
array([[9, 7, 6]]),
array([[1, 6, 3]])]
In [50]: np.stack(some_df['A'].to_list()).shape
Out[50]: (4, 1, 3)
To take an input:
def chunks(l, n):
return [l[i:i+n] for i in range(0, len(l), n)]
mylist = [1,2,3,4,5,6,7,8]
while 1:
try:
size = int(raw_input('What size? ')) # Or input() if python 3.x
break
except ValueError:
print "Numbers only please"
print chunks(yourlist, size)
Prints:
[[1, 2], [3, 4], [5, 6], [7, 8]] # Assuming 2 was the input
Or even:
>>> zip(*[iter(l)]*size) # Assuming 2 was the input
[(1, 2), (3, 4), (5, 6), (7, 8)]
The point is that say a vector can be seen either as
You can add dimensions using
[:, np.newaxis]syntax or drop dimensions usingnp.squeeze: