How to refine a mesh in python quickly - python

I have a numpy array([1.0, 2.0, 3.0]), which is actually a mesh in 1 dimension in my problem. What I want to do is to refine the mesh to get this: array([0.8, 0.9, 1, 1.1, 1.2, 1.8, 1.9, 2, 2.1, 2.2, 2.8, 2.9, 3, 3.1, 3.2,]).
The actual array is very large and this procedure costs a lot of time. How to do this quickly (maybe vectorize) in python?

Here's a vectorized approach -
(a[:,None] + np.arange(-0.2,0.3,0.1)).ravel() # a is input array
Sample run -
In [15]: a = np.array([1.0, 2.0, 3.0]) # Input array
In [16]: (a[:,None] + np.arange(-0.2,0.3,0.1)).ravel()
Out[16]:
array([ 0.8, 0.9, 1. , 1.1, 1.2, 1.8, 1.9, 2. , 2.1, 2.2, 2.8,
2.9, 3. , 3.1, 3.2])

Here are a few options(python 3):
Option 1:
np.array([j for i in arr for j in np.arange(i - 0.2, i + 0.25, 0.1)])
# array([ 0.8, 0.9, 1. , 1.1, 1.2, 1.8, 1.9, 2. , 2.1, 2.2, 2.8,
# 2.9, 3. , 3.1, 3.2])
Option 2:
np.array([j for x, y in zip(arr - 0.2, arr + 0.25) for j in np.arange(x,y,0.1)])
# array([ 0.8, 0.9, 1. , 1.1, 1.2, 1.8, 1.9, 2. , 2.1, 2.2, 2.8,
# 2.9, 3. , 3.1, 3.2])
Option 3:
np.array([arr + i for i in np.arange(-0.2, 0.25, 0.1)]).T.ravel()
# array([ 0.8, 0.9, 1. , 1.1, 1.2, 1.8, 1.9, 2. , 2.1, 2.2, 2.8,
# 2.9, 3. , 3.1, 3.2])
Timing on a larger array:
arr = np.arange(100000)
arr
# array([ 0, 1, 2, ..., 99997, 99998, 99999])
%timeit np.array([j for i in arr for j in np.arange(i-0.2, i+0.25, 0.1)])
# 1 loop, best of 3: 615 ms per loop
%timeit np.array([j for x, y in zip(arr - 0.2, arr + 0.25) for j in np.arange(x,y,0.1)])
# 1 loop, best of 3: 250 ms per loop
%timeit np.array([arr + i for i in np.arange(-0.2, 0.25, 0.1)]).T.ravel()
# 100 loops, best of 3: 1.93 ms per loop

Related

How to convert np.array into pd.DataFrame

I have loaded the 'load_iris' toy dataset in the Scikit learn library.
{'data': array([[5.1, 3.5, 1.4, 0.2],
[4.9, 3. , 1.4, 0.2],
[4.7, 3.2, 1.3, 0.2],
[4.6, 3.1, 1.5, 0.2],
[5. , 3.6, 1.4, 0.2],
[5.4, 3.9, 1.7, 0.4],
[4.6, 3.4, 1.4, 0.3],
[5. , 3.4, 1.5, 0.2],
[4.4, 2.9, 1.4, 0.2],
[4.9, 3.1, 1.5, 0.1],
[5.4, 3.7, 1.5, 0.2],
[4.8, 3.4, 1.6, 0.2],
[4.8, 3. , 1.4, 0.1],
[4.3, 3. , 1.1, 0.1],
[5.8, 4. , 1.2, 0.2],
[5.7, 4.4, 1.5, 0.4],
[5.4, 3.9, 1.3, 0.4],
[5.1, 3.5, 1.4, 0.3],
[5.7, 3.8, 1.7, 0.3],
[5.1, 3.8, 1.5, 0.3],
[5.4, 3.4, 1.7, 0.2],
[5.1, 3.7, 1.5, 0.4],
[4.6, 3.6, 1. , 0.2],
[5.1, 3.3, 1.7, 0.5],
[4.8, 3.4, 1.9, 0.2],
[5. , 3. , 1.6, 0.2],
[5. , 3.4, 1.6, 0.4],
[5.2, 3.5, 1.5, 0.2],
[5.2, 3.4, 1.4, 0.2],
[4.7, 3.2, 1.6, 0.2],
[4.8, 3.1, 1.6, 0.2],
[5.4, 3.4, 1.5, 0.4],
[5.2, 4.1, 1.5, 0.1],
[5.5, 4.2, 1.4, 0.2],
[4.9, 3.1, 1.5, 0.2],
[5. , 3.2, 1.2, 0.2],
[5.5, 3.5, 1.3, 0.2],
[4.9, 3.6, 1.4, 0.1],
[4.4, 3. , 1.3, 0.2],
[5.1, 3.4, 1.5, 0.2],
[5. , 3.5, 1.3, 0.3],
[4.5, 2.3, 1.3, 0.3],
[4.4, 3.2, 1.3, 0.2],
[5. , 3.5, 1.6, 0.6],
[5.1, 3.8, 1.9, 0.4],
[4.8, 3. , 1.4, 0.3],
[5.1, 3.8, 1.6, 0.2],
[4.6, 3.2, 1.4, 0.2],
[5.3, 3.7, 1.5, 0.2],
[5. , 3.3, 1.4, 0.2],
[7. , 3.2, 4.7, 1.4],
[6.4, 3.2, 4.5, 1.5],
[6.9, 3.1, 4.9, 1.5],
[5.5, 2.3, 4. , 1.3],
[6.5, 2.8, 4.6, 1.5],
[5.7, 2.8, 4.5, 1.3],
[6.3, 3.3, 4.7, 1.6],
[4.9, 2.4, 3.3, 1. ],
[6.6, 2.9, 4.6, 1.3],
[5.2, 2.7, 3.9, 1.4],
[5. , 2. , 3.5, 1. ],
[5.9, 3. , 4.2, 1.5],
[6. , 2.2, 4. , 1. ],
[6.1, 2.9, 4.7, 1.4],
[5.6, 2.9, 3.6, 1.3],
[6.7, 3.1, 4.4, 1.4],
[5.6, 3. , 4.5, 1.5],
[5.8, 2.7, 4.1, 1. ],
[6.2, 2.2, 4.5, 1.5],
[5.6, 2.5, 3.9, 1.1],
[5.9, 3.2, 4.8, 1.8],
[6.1, 2.8, 4. , 1.3],
[6.3, 2.5, 4.9, 1.5],
[6.1, 2.8, 4.7, 1.2],
[6.4, 2.9, 4.3, 1.3],
[6.6, 3. , 4.4, 1.4],
[6.8, 2.8, 4.8, 1.4],
[6.7, 3. , 5. , 1.7],
[6. , 2.9, 4.5, 1.5],
[5.7, 2.6, 3.5, 1. ],
[5.5, 2.4, 3.8, 1.1],
[5.5, 2.4, 3.7, 1. ],
[5.8, 2.7, 3.9, 1.2],
[6. , 2.7, 5.1, 1.6],
[5.4, 3. , 4.5, 1.5],
[6. , 3.4, 4.5, 1.6],
[6.7, 3.1, 4.7, 1.5],
[6.3, 2.3, 4.4, 1.3],
[5.6, 3. , 4.1, 1.3],
[5.5, 2.5, 4. , 1.3],
[5.5, 2.6, 4.4, 1.2],
[6.1, 3. , 4.6, 1.4],
[5.8, 2.6, 4. , 1.2],
[5. , 2.3, 3.3, 1. ],
[5.6, 2.7, 4.2, 1.3],
[5.7, 3. , 4.2, 1.2],
[5.7, 2.9, 4.2, 1.3],
[6.2, 2.9, 4.3, 1.3],
[5.1, 2.5, 3. , 1.1],
[5.7, 2.8, 4.1, 1.3],
[6.3, 3.3, 6. , 2.5],
[5.8, 2.7, 5.1, 1.9],
[7.1, 3. , 5.9, 2.1],
[6.3, 2.9, 5.6, 1.8],
[6.5, 3. , 5.8, 2.2],
[7.6, 3. , 6.6, 2.1],
[4.9, 2.5, 4.5, 1.7],
[7.3, 2.9, 6.3, 1.8],
[6.7, 2.5, 5.8, 1.8],
[7.2, 3.6, 6.1, 2.5],
[6.5, 3.2, 5.1, 2. ],
[6.4, 2.7, 5.3, 1.9],
[6.8, 3. , 5.5, 2.1],
[5.7, 2.5, 5. , 2. ],
[5.8, 2.8, 5.1, 2.4],
[6.4, 3.2, 5.3, 2.3],
[6.5, 3. , 5.5, 1.8],
[7.7, 3.8, 6.7, 2.2],
[7.7, 2.6, 6.9, 2.3],
[6. , 2.2, 5. , 1.5],
[6.9, 3.2, 5.7, 2.3],
[5.6, 2.8, 4.9, 2. ],
[7.7, 2.8, 6.7, 2. ],
[6.3, 2.7, 4.9, 1.8],
[6.7, 3.3, 5.7, 2.1],
[7.2, 3.2, 6. , 1.8],
[6.2, 2.8, 4.8, 1.8],
[6.1, 3. , 4.9, 1.8],
[6.4, 2.8, 5.6, 2.1],
[7.2, 3. , 5.8, 1.6],
[7.4, 2.8, 6.1, 1.9],
[7.9, 3.8, 6.4, 2. ],
[6.4, 2.8, 5.6, 2.2],
[6.3, 2.8, 5.1, 1.5],
[6.1, 2.6, 5.6, 1.4],
[7.7, 3. , 6.1, 2.3],
[6.3, 3.4, 5.6, 2.4],
[6.4, 3.1, 5.5, 1.8],
[6. , 3. , 4.8, 1.8],
[6.9, 3.1, 5.4, 2.1],
[6.7, 3.1, 5.6, 2.4],
[6.9, 3.1, 5.1, 2.3],
[5.8, 2.7, 5.1, 1.9],
[6.8, 3.2, 5.9, 2.3],
[6.7, 3.3, 5.7, 2.5],
[6.7, 3. , 5.2, 2.3],
[6.3, 2.5, 5. , 1.9],
[6.5, 3. , 5.2, 2. ],
[6.2, 3.4, 5.4, 2.3],
[5.9, 3. , 5.1, 1.8]]),
'target': array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]),
'frame': None,
'target_names': array(['setosa', 'versicolor', 'virginica'], dtype='<U10'),
'DESCR': '.. _iris_dataset:\n\nIris plants dataset\n--------------------\n\n**Data Set Characteristics:**\n\n :Number of Instances: 150 (50 in each of three classes)\n :Number of Attributes: 4 numeric, predictive attributes and the class\n :Attribute Information:\n - sepal length in cm\n - sepal width in cm\n - petal length in cm\n - petal width in cm\n - class:\n - Iris-Setosa\n - Iris-Versicolour\n - Iris-Virginica\n \n :Summary Statistics:\n\n ============== ==== ==== ======= ===== ====================\n Min Max Mean SD Class Correlation\n ============== ==== ==== ======= ===== ====================\n sepal length: 4.3 7.9 5.84 0.83 0.7826\n sepal width: 2.0 4.4 3.05 0.43 -0.4194\n petal length: 1.0 6.9 3.76 1.76 0.9490 (high!)\n petal width: 0.1 2.5 1.20 0.76 0.9565 (high!)\n ============== ==== ==== ======= ===== ====================\n\n :Missing Attribute Values: None\n :Class Distribution: 33.3% for each of 3 classes.\n :Creator: R.A. Fisher\n :Donor: Michael Marshall (MARSHALL%PLU#io.arc.nasa.gov)\n :Date: July, 1988\n\nThe famous Iris database, first used by Sir R.A. Fisher. The dataset is taken\nfrom Fisher\'s paper. Note that it\'s the same as in R, but not as in the UCI\nMachine Learning Repository, which has two wrong data points.\n\nThis is perhaps the best known database to be found in the\npattern recognition literature. Fisher\'s paper is a classic in the field and\nis referenced frequently to this day. (See Duda & Hart, for example.) The\ndata set contains 3 classes of 50 instances each, where each class refers to a\ntype of iris plant. One class is linearly separable from the other 2; the\nlatter are NOT linearly separable from each other.\n\n.. topic:: References\n\n - Fisher, R.A. "The use of multiple measurements in taxonomic problems"\n Annual Eugenics, 7, Part II, 179-188 (1936); also in "Contributions to\n Mathematical Statistics" (John Wiley, NY, 1950).\n - Duda, R.O., & Hart, P.E. (1973) Pattern Classification and Scene Analysis.\n (Q327.D83) John Wiley & Sons. ISBN 0-471-22361-1. See page 218.\n - Dasarathy, B.V. (1980) "Nosing Around the Neighborhood: A New System\n Structure and Classification Rule for Recognition in Partially Exposed\n Environments". IEEE Transactions on Pattern Analysis and Machine\n Intelligence, Vol. PAMI-2, No. 1, 67-71.\n - Gates, G.W. (1972) "The Reduced Nearest Neighbor Rule". IEEE Transactions\n on Information Theory, May 1972, 431-433.\n - See also: 1988 MLC Proceedings, 54-64. Cheeseman et al"s AUTOCLASS II\n conceptual clustering system finds 3 classes in the data.\n - Many, many more ...',
'feature_names': ['sepal length (cm)',
'sepal width (cm)',
'petal length (cm)',
'petal width (cm)'],
'filename': 'iris.csv',
'data_module': 'sklearn.datasets.data'}
I wish to convert this dataset, which is in array form into a data frame but am unable to do so with the following command, which return the first 4 columns completely filled with Nan
y = pd.DataFrame(datasets.load_iris(),columns = ['sepal length (cm)','sepal width (cm)','petal length (cm)','petal width (cm)','target'])
The command gives the following table, which is not correct
sepal length (cm) sepal width (cm) petal length (cm) petal width (cm) target
0 NaN NaN NaN NaN 0
1 NaN NaN NaN NaN 0
2 NaN NaN NaN NaN 0
3 NaN NaN NaN NaN 0
4 NaN NaN NaN NaN 0
... ... ... ... ... ...
145 NaN NaN NaN NaN 2
146 NaN NaN NaN NaN 2
147 NaN NaN NaN NaN 2
148 NaN NaN NaN NaN 2
149 NaN NaN NaN NaN 2
How to do it?
How to get data correctly converted from np.array into pd.DataFrame
Use the as_frame=True option:
df = datasets.load_iris(as_frame=True)['data']
output:
sepal length (cm) sepal width (cm) petal length (cm) petal width (cm)
0 5.1 3.5 1.4 0.2
1 4.9 3.0 1.4 0.2
2 4.7 3.2 1.3 0.2
3 4.6 3.1 1.5 0.2
4 5.0 3.6 1.4 0.2
.. ... ... ... ...
145 6.7 3.0 5.2 2.3
146 6.3 2.5 5.0 1.9
147 6.5 3.0 5.2 2.0
148 6.2 3.4 5.4 2.3
149 5.9 3.0 5.1 1.8
[150 rows x 4 columns]
If you also want the target:
iris = datasets.load_iris(as_frame=True)
df = iris['data']
df['target'] = iris['target']
output:
sepal length (cm) sepal width (cm) petal length (cm) petal width (cm) target
0 5.1 3.5 1.4 0.2 0
1 4.9 3.0 1.4 0.2 0
2 4.7 3.2 1.3 0.2 0
3 4.6 3.1 1.5 0.2 0
4 5.0 3.6 1.4 0.2 0
.. ... ... ... ... ...
145 6.7 3.0 5.2 2.3 2
146 6.3 2.5 5.0 1.9 2
147 6.5 3.0 5.2 2.0 2
148 6.2 3.4 5.4 2.3 2
149 5.9 3.0 5.1 1.8 2

How to use a different colormap for different rows of a heatmap

I am trying to change 1 row in my heatmap to a different color
here is the dataset:
m = np.array([[ 0.7, 1.4, 0.2, 1.5, 1.7, 1.2, 1.5, 2.5],
[ 1.1, 2.5, 0.4, 1.7, 2. , 2.4, 2. , 3.2],
[ 0.9, 4.4, 0.7, 2.3, 1.6, 2.3, 2.6, 3.3],
[ 0.8, 2.1, 0.2, 1.8, 2.3, 1.9, 2. , 2.9],
[ 0.9, 1.3, 0.8, 2.2, 1.8, 2.2, 1.7, 2.8],
[ 0.7, 0.9, 0.4, 1.8, 1.4, 2.1, 1.7, 2.9],
[ 1.2, 0.9, 0.4, 2.1, 1.3, 1.2, 1.9, 2.4],
[ 6.3, 13.5, 3.1, 13.4, 12.1, 13.3, 13.4, 20. ]])
data = pd.DataFrame(data = m)
Right now I am using seaborn heatmap, I can only create something like this:
cmap = sns.diverging_palette(240, 10, as_cmap = True)
sns.heatmap(data, annot = True, cmap = "Reds")
plt.show
I hope to change the color scheme of the last row, here is what I want to achieve (I did this in Excel):
Is it possible I achieve this in Python with seaborn heatmap? Thank you!
You can split in two, mask the unwanted parts, and plot separately:
# Reds
data1 = data.copy()
data1.loc[7] = float('nan')
ax = sns.heatmap(data1, annot=True, cmap="Reds")
# Greens
data2 = data.copy()
data2.loc[:6] = float('nan')
sns.heatmap(data2, annot=True, cmap="Greens")
output:
NB. you need to adapt the loc[…] parameter to your actual index names

is there a parameter to set the precision for numpy.linspace?

I am trying to check if a numpy array contains a specific value:
>>> x = np.linspace(-5,5,101)
>>> x
array([-5. , -4.9, -4.8, -4.7, -4.6, -4.5, -4.4, -4.3, -4.2, -4.1, -4. ,
-3.9, -3.8, -3.7, -3.6, -3.5, -3.4, -3.3, -3.2, -3.1, -3. , -2.9,
-2.8, -2.7, -2.6, -2.5, -2.4, -2.3, -2.2, -2.1, -2. , -1.9, -1.8,
-1.7, -1.6, -1.5, -1.4, -1.3, -1.2, -1.1, -1. , -0.9, -0.8, -0.7,
-0.6, -0.5, -0.4, -0.3, -0.2, -0.1, 0. , 0.1, 0.2, 0.3, 0.4,
0.5, 0.6, 0.7, 0.8, 0.9, 1. , 1.1, 1.2, 1.3, 1.4, 1.5,
1.6, 1.7, 1.8, 1.9, 2. , 2.1, 2.2, 2.3, 2.4, 2.5, 2.6,
2.7, 2.8, 2.9, 3. , 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7,
3.8, 3.9, 4. , 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8,
4.9, 5. ])
>>> -5. in x
True
>>> a = 0.2
>>> a
0.2
>>> a in x
False
I assigned a constant to variable a. It seems that the precision of a is not compatible with the elements in the numpy array generated by np.linspace().
I've searched the docs, but didn't find anything about this.
This is not a question of the precision of np.linspace, but rather of the type of the elements in the generated array.
np.linspace generates elements which, conceptually, equally divide the input range between them. However, these elements are then stored as floating point numbers with limited precision, which makes the generation process itself appear to lack precision.
By passing the dtype argument to np.linspace, you can specify the precision of the floating point type used to store its result, which can increase the apparent precision of the generation process.
Nevertheless, you should not use the equality operator to compare floating point numbers. Instead, use np.isclose in conjunction with np.ndarray.any, or some equivalent:
>>> floats_64 = np.linspace(-5, 5, 101, dtype='float64')
>>> floats_128 = np.linspace(-5, 5, 101, dtype='float128')
>>> print(0.2 in floats_64)
False
>>> print(floats_64[52])
0.20000000000000018
>>> print(np.isclose(0.2, floats_64).any()) # check if any element in floats_64 is close to 0.2
True
>>> print(0.2 in floats_128)
False
>>> print(floats_128[52])
0.20000000000000017764
>>> print(np.isclose(0.2, floats_128).any()) # check if any element in floats_128 is close to 0.2
True

How to print a value to a new array if it within a bound of previous value in that array in Python/Numpy

If I have an array:
StartArray=np.array([1, 2, 3, 1.4, 1.2, 0.6, 1.8, 1.5, 1.9, 2.2, 3, 4 ,2.3])
I would like to loop through this array starting with StartArray[0] and only keep values that are within +/- .5 of the last kept value to yield:
EndArray=[1, 1.4, 1.2, 1.5, 1.9, 2.2, 2.3]
This is what I have tried so far and the results don't make sense
StartArray=np.array([1, 2, 3, 1.4, 1.2, 0.6, 1.8, 1.5, 1.9, 2.2, 3, 4 ,2.3])
EndArray=np.empty_like(StartArray)
EndArray[0]=StartArray[0]
for i in range(len(StartArray)-1):
if EndArray[i]+.5>StartArray[i+1]>EndArray[i]-.5:
EndArray[i+1]=StartArray[i+1]
Out:
array([ 1. , 0.22559146, 0.13015365, 5.24910493, 0.63804761,
0.6 , 1.73143364, 1.5 , 1.9 , 2.2 ,
6.82525036, 0.61641556, 6.82325036])
List is the good structure for this job:
StartArray=np.array([1, 2, 3, 1.4, 1.2, 0.6, 1.8, 1.5, 1.9, 2.2, 3, 4 ,2.3])
ref=StartArray[0]
End=[]
for x in StartArray:
if abs(x- ref)<.5:
End.append(x)
ref=x
print(np.array(End))
[ 1. 1.4 1.2 1.5 1.9 2.2 2.3]
There are multiple problems with your approach. First, you're initializing EndArray to be the same size as StartArray, but that's not what you want your desired output to be. Instead, initialize EndArray to be an empty list and append values as your loop through StartArray. Secondly, you want the output values to be within 0.5 of the last kept value, so you need to keep track of this.
Adapting your code:
StartArray=np.array([1, 2, 3, 1.4, 1.2, 0.6, 1.8, 1.5, 1.9, 2.2, 3, 4 ,2.3])
EndArray=[]
last_kept = StartArray[0]
EndArray.append(last_kept)
for i in range(len(StartArray)-1):
if np.abs(StartArray[i+1] - last_kept) < 0.5:
last_kept = StartArray[i+1]
EndArray.append(last_kept)
# convert back to numpy array
EndArray = np.array(EndArray)

Python removing all negative values in array

What is the most efficient way to remove negative elements in an array? I have tried numpy.delete and Remove all specific value from array and code of the form x[x != i].
For:
import numpy as np
x = np.array([-2, -1.4, -1.1, 0, 1.2, 2.2, 3.1, 4.4, 8.3, 9.9, 10, 14, 16.2])
I want to end up with an array:
[0, 1.2, 2.2, 3.1, 4.4, 8.3, 9.9, 10, 14, 16.2]
In [2]: x[x >= 0]
Out[2]: array([ 0. , 1.2, 2.2, 3.1, 4.4, 8.3, 9.9, 10. , 14. , 16.2])
If performance is important, you could take advantage of the fact that your np.array is sorted and use numpy.searchsorted
For example:
In [8]: x[np.searchsorted(x, 0) :]
Out[8]: array([ 0. , 1.2, 2.2, 3.1, 4.4, 8.3, 9.9, 10. , 14. , 16.2])
In [9]: %timeit x[np.searchsorted(x, 0) :]
1000000 loops, best of 3: 1.47 us per loop
In [10]: %timeit x[x >= 0]
100000 loops, best of 3: 4.5 us per loop
The difference in performance will increase as the size of the array increases because np.searchsorted does a binary search that is O(log n) vs. O(n) linear search that x >= 0 is doing.
In [11]: x = np.arange(-1000, 1000)
In [12]: %timeit x[np.searchsorted(x, 0) :]
1000000 loops, best of 3: 1.61 us per loop
In [13]: %timeit x[x >= 0]
100000 loops, best of 3: 9.87 us per loop
In numpy:
b = array[array>=0]
Example:
>>> import numpy as np
>>> arr = np.array([-2, -1.4, -1.1, 0, 1.2, 2.2, 3.1, 4.4, 8.3, 9.9, 10, 14, 16.2])
>>> arr = arr[arr>=0]
>>> arr
array([ 0. , 1.2, 2.2, 3.1, 4.4, 8.3, 9.9, 10. , 14. , 16.2])
There's probably a cool way to do this is numpy because numpy is magic to me, but:
x = np.array( [ num for num in x if num >= 0 ] )

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