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How to find list intersection?
...y about duplicates then you can use set intersection:
>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> list(set(a) & set(b))
[1, 3, 5]
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Phonegap Cordova installation Windows
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frigonfrigon
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Fast Bitmap Blur For Android SDK
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answered Jan 15 '10 at 2:03
LukeLuke
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How to calculate the number of occurrence of a given character in each row of a column of strings?
...a")
q.data
# number string number.of.a
#1 1 greatgreat 2
#2 2 magic 1
#3 3 not 0
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Reverse Range in Swift
...sed() method on a range
for i in (1...5).reversed() { print(i) } // 5 4 3 2 1
Or stride(from:through:by:) method
for i in stride(from:5,through:1,by:-1) { print(i) } // 5 4 3 2 1
stide(from:to:by:) is similar but excludes the last value
for i in stride(from:5,to:0,by:-1) { print(i) } // 5 4 3...
Could not calculate build plan: Plugin org.apache.maven.plugins:maven-resources-plugin:2.5 or one of
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How to insert element into arrays at specific position?
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How can I check if a program exists from a Bash script?
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dropping infinite values from dataframes in pandas?
... dropna:
df.replace([np.inf, -np.inf], np.nan).dropna(subset=["col1", "col2"], how="all")
For example:
In [11]: df = pd.DataFrame([1, 2, np.inf, -np.inf])
In [12]: df.replace([np.inf, -np.inf], np.nan)
Out[12]:
0
0 1
1 2
2 NaN
3 NaN
The same method would work for a Series.
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numpy: most efficient frequency counts for unique values in an array
...ence/generated/numpy.bincount.html
import numpy as np
x = np.array([1,1,1,2,2,2,5,25,1,1])
y = np.bincount(x)
ii = np.nonzero(y)[0]
And then:
zip(ii,y[ii])
# [(1, 5), (2, 3), (5, 1), (25, 1)]
or:
np.vstack((ii,y[ii])).T
# array([[ 1, 5],
[ 2, 3],
[ 5, 1],
[25, ...