If i need to ask a condition on every element of a numpy.ndarray of integers, do I have to use a for loop
for i in range(n):
if a[i] == 0:
a[i] = 1
or can I ask the question using [:] syntax
if a[:] == 0:
#...
I know the previous is wrong, but is there any way of doing something similar?
You can use the all builtin function to accomplish what your asking:
all(i == 0 for i in a)
Example:
>>> a = [1, 0, 0, 2, 3, 0]
>>> all(i == 0 for i in a)
False
Note however that behinds the scenes, all still uses a for loop. It's just implemented in C:
for (;;) {
item = iternext(it);
if (item == NULL)
break;
cmp = PyObject_IsTrue(item);
Py_DECREF(item);
if (cmp < 0) {
Py_DECREF(it);
return NULL;
}
if (cmp == 0) {
Py_DECREF(it);
Py_RETURN_FALSE;
}
EDIT: Given your most recent edits, what you probably want instead is to use a list comprehension with the ternary operator:
[1 if i == 0 else i for i in a]
Example:
>>> a = [1, 0, 0, 2, 3, 0]
>>> [1 if i == 0 else i for i in a]
[1, 1, 1, 2, 3, 1]
if to come later.
For testing a condition on every element of a numpy.ndarray at once, as the title could suggest:
use numpy's np.all for that:
if np.all(a == 0):
# ...
Despite it is not lazy, np.all is vectorized and very fast
# arrays of zeros
>>> a = np.zeros((1000000))
>>> %timeit np.all(a == 0) # vectorized, very fast
10000 loops, best of 3: 34.5 µs per loop
>>>%timeit all(i == 0 for i in a) # not vectorized...
100 loops, best of 3: 19.3 ms per loop
# arrays of non-zeros
>>> b = np.ones((1000000))
>>> %timeit np.all(b == 0) # not lazy, iterates through all array
1000 loops, best of 3: 498 µs per loop
>>> %timeit all(i == 0 for i in b) # lazy, return false at first 1
1000000 loops, best of 3: 561 ns per loop
# n-D arrays of zeros
>>> c = a.reshape((100, 1000)) # 2D array
>>> %timeit np.all(c == 0)
10000 loops, best of 3: 34.7 µs per loop # works on n-dim arrays
>>> %timeit all(i == 0 for i in c) # wors for a 1D arrays only
...
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
For testing a condition on every element of a numpy.ndarray iteratively:
for i in range(n):
if a[i] == 0:
a[i] = 1
can be replaced by np.where
a = np.where(a == 0, 1, a) # set value '1' where condition is met
EDIT: precisions according to the OP's comments
Assuming a is your array, and you want to change values of a that are greater than 1 to be equal to 1:
a[a > 1] = 1
This works because the expression a > 1 creates a mask array, and when a mask array is used as an index (which it is here), the operation only applies on the True indices.
a[a == 0] = 1. All the other answers are unnecessarily complicated.if you need not just check, but map all 0 --> 1, use map:
map(lambda x: 1 if x==0 else x, a)
#...supposed to do. Do you want to call a code fragment for every element?a? Python array or numpy array?