how can i extract all the odd number from a numpy array?
try:
import numpy as np
a = np.array([1,2,3,4,5,6,6,7,7,8,9])
a[a % 2 == 1]
Out[13]: array([1, 3, 5, 7, 7, 9])
b = np.where(a%2)
print(f'Array with Odd Numbers: {a[b]}')
Here a is the array containing all the numbers.
And the Output of Odd Numbers from that array is given as a[b]
Related
I have a 3-D NumPy array with shape (100, 50, 20). I was trying to slice the third dimension of the array by using the index, e.g., from 1 to 6 and from 8 to 10.
I tried the following code, but it kept reporting a syntax error.
newarr [:,:,1:10] = oldarr[:,:,[1:7,8:11]]
You can use np.r_ to concatenate slice objects:
newarr [:,:,1:10] = oldarr[:,:,np.r_[1:7,8:11]]
Example:
np.r_[1:4,6:8]
array([1, 2, 3, 6, 7])
I have a DataFrame with two pandas Series as follow:
value accepted_values
0 1 [1, 2, 3, 4]
1 2 [5, 6, 7, 8]
I would like to efficiently check if the value is in accepted_values using pandas methods.
I already know I can do something like the following, but I'm interested in a faster approach if there is one (took around 27 seconds on 1 million rows DataFrame)
import pandas as pd
df = pd.DataFrame({"value":[1, 2], "accepted_values": [[1,2,3,4], [5, 6, 7, 8]]})
def check_first_in_second(values: pd.Series):
return values[0] in values[1]
are_in_accepted_values = df[["value", "accepted_values"]].apply(
check_first_in_second, axis=1
)
if not are_in_accepted_values.all():
raise AssertionError("Not all value in accepted_values")
I think if create DataFrame with list column you can compare by DataFrame.eq and test if match at least one value per row by DataFrame.any:
df1 = pd.DataFrame(df["accepted_values"].tolist(), index=df.index)
are_in_accepted_values = df1.eq(df["value"]).any(axis=1).all()
Another idea:
are_in_accepted_values = all(v in a for v, a in df[["value", "accepted_values"]].to_numpy())
I found a little optimisation to your second idea. Using a bit more numpy than pandas makes it faster (more than 3x, tested with time.perf_counter()).
values = df["value"].values
accepted_values = df["accepted_values"].values
are_in_accepted_values = all(s in e for s, e in np.column_stack([values, accepted_values]))
Is there a way to find out the highest individual numbers give a list of Numpy arrays?
e.g.
import numpy as np
a = np.array([10, 2])
b = np.array([3, 4])
c = np.array([5, 6])
The output will be a Numpy array with the following form:
np.array([10, 6])
You can stack all arrays together and get the max afterwards:
np.stack((a, b, c)).max(0)
# array([10, 6])
I want to get all unique combinations of a numpy.array vector (or a pandas.Series). I used itertools.combinations but it's very slow. For an array of size (1000,) it takes many hours. Here is my code using itertools (actually I use combination differences):
def a(array):
temp = pd.Series([])
for i in itertools.combinations(array, 2):
temp = temp.append(pd.Series(np.abs(i[0]-i[1])))
temp.index=range(len(temp))
return temp
As you see there is no repetition!!
The sklearn.utils.extmath.cartesian is really fast and good but it provides repetitions which I do not want! I need help rewriting above function without using itertools and much more speed for large vectors.
You could take the upper triangular part of a matrix formed on the Cartesian product with the binary operation (here subtraction, as in your example):
import numpy as np
n = 3
a = np.random.randn(n)
print(a)
print(a - a[:, np.newaxis])
print((a - a[:, np.newaxis])[np.triu_indices(n, 1)])
gives
[ 0.04248369 -0.80162228 -0.44504522]
[[ 0. -0.84410597 -0.48752891]
[ 0.84410597 0. 0.35657707]
[ 0.48752891 -0.35657707 0. ]]
[-0.84410597 -0.48752891 0.35657707]
with n=1000 (and output piped to /dev/null) this runs in 0.131s
on my relatively modest laptop.
For a random array of ints:
import numpy as np
import pandas as pd
import itertools as it
b = np.random.randint(0, 8, ((6,)))
# array([7, 0, 6, 7, 1, 5])
pd.Series(list(it.combinations(np.unique(b), 2)))
it returns:
0 (0, 1)
1 (0, 5)
2 (0, 6)
3 (0, 7)
4 (1, 5)
5 (1, 6)
6 (1, 7)
7 (5, 6)
8 (5, 7)
9 (6, 7)
dtype: object
How to achieve this?
I have a numpy array containing:
[1, 2, 3]
I want to create an array containing:
[8, 1, 2, 3]
That is, I want to add an element on as the first element of the array.
Ref:Add single element to array in numpy
The most basic operation is concatenate:
x=np.array([1,2,3])
np.concatenate([[8],x])
# array([8, 1, 2, 3])
np.r_ and np.insert make use of this. Even if they are more convenient to remember, or use in more complex cases, you should be familiar with concatenate.
Use numpy.insert(). The docs are here: http://docs.scipy.org/doc/numpy/reference/generated/numpy.insert.html#numpy.insert
You can also use numpy's np.r_, a short-cut for concatenation along the first axis:
>>> import numpy as np
>>> a = np.array([1, 2, 3])
>>> b = np.r_[8, a]
>>> b
array([8, 1, 2, 3])