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What's the standard way to work with dates and times in Scala? Should I use Java types or there are
...cala_time.time.Imports._
DateTime.now // returns org.joda.time.DateTime = 2009-04-27T13:25:42.659-07:00
DateTime.now.hour(2).minute(45).second(10) // returns org.joda.time.DateTime = 2009-04-27T02:45:10.313-07:00
DateTime.now + 2.months // returns org.joda.time.DateTime = 2009-06-27T13:25:59.195-...
How can I include a YAML file inside another?
...
jameshfisherjameshfisher
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Grep characters before and after match?
...
3 characters before and 4 characters after
$> echo "some123_string_and_another" | grep -o -P '.{0,3}string.{0,4}'
23_string_and
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Encode URL in JavaScript?
...
2847
Check out the built-in function encodeURIComponent(str) and encodeURI(str).
In your case, thi...
Compare two DataFrames and output their differences side-by-side
...lar to Constantine, you can get the boolean of which rows are empty*:
In [21]: ne = (df1 != df2).any(1)
In [22]: ne
Out[22]:
0 False
1 True
2 True
dtype: bool
Then we can see which entries have changed:
In [23]: ne_stacked = (df1 != df2).stack()
In [24]: changed = ne_stacked[ne_stac...
jQuery get value of select onChange
...alue );
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<select>
<option value="1">One</option>
<option value="2">Two</option>
</select>
You can also reference with onchange event-
...
How do I read a text file of about 2 GB? [duplicate]
I have a .txt file whose memory is more than 2 GB. The problem is I cannot open it with Notepad, Notepad++ or any other editor programs.
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Logical XOR operator in C++?
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answered Oct 20 '09 at 19:03
Greg HewgillGreg Hewgill
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How to get the first column of a pandas DataFrame as a Series?
...>>> import pandas as pd
>>> df = pd.DataFrame({'x' : [1, 2, 3, 4], 'y' : [4, 5, 6, 7]})
>>> df
x y
0 1 4
1 2 5
2 3 6
3 4 7
>>> s = df.ix[:,0]
>>> type(s)
<class 'pandas.core.series.Series'>
>>>
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Peak-finding algorithm for Python/SciPy
... matplotlib.pyplot as plt
from scipy.signal import find_peaks
x = np.sin(2*np.pi*(2**np.linspace(2,10,1000))*np.arange(1000)/48000) + np.random.normal(0, 1, 1000) * 0.15
peaks, _ = find_peaks(x, distance=20)
peaks2, _ = find_peaks(x, prominence=1) # BEST!
peaks3, _ = find_peaks(x, width=20)
p...