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Print a list in reverse order with range()?
...rameters. range(start, stop, step)
For example, to generate a list [5,4,3,2,1,0], you can use the following:
range(5, -1, -1)
It may be less intuitive but as the comments mention, this is more efficient and the right usage of range for reversed list.
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How to iterate over a JavaScript object?
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edited Nov 3 '19 at 12:50
Beachhouse
4,46722 gold badges2222 silver badges3434 bronze badges
answer...
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...
How can I include a YAML file inside another?
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jameshfisherjameshfisher
24.3k2020 gold badges8484 silver badges137137 bronze badges
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Grep characters before and after match?
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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?
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2847
Check out the built-in function encodeURIComponent(str) and encodeURI(str).
In your case, thi...
Logical XOR operator in C++?
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answered Oct 20 '09 at 19:03
Greg HewgillGreg Hewgill
783k167167 gold badges10841084 silver badges12221222 bronze badges
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C++常用排序算法汇总 - C/C++ - 清泛网 - 专注C/C++及内核技术
...****************************
* 简单选择排序
* 稳定排序，O{n^2} ~ O{n^2}
* 从首位开始，循环一次找出一个比首位小的值，交换
*
* https://www.tsingfun.com
************************************/
#include<stdio.h>
#include<stdlib.h>
/*
第一种形式的选...
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-
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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...