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Accessing last x characters of a string in Bash
I found out that with ${string:0:3} one can access the first 3 characters of a string. Is there a equivalently easy method to access the last three characters?
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How can I obtain the element-wise logical NOT of a pandas Series?
... True, False, True])
In [8]: ~s
Out[8]:
0 False
1 False
2 True
3 False
dtype: bool
Using Python2.7, NumPy 1.8.0, Pandas 0.13.1:
In [119]: s = pd.Series([True, True, False, True]*10000)
In [10]: %timeit np.invert(s)
10000 loops, best of 3: 91.8 µs per loop
In [11]: %timeit ~s
10...
What are the correct version numbers for C#?
...numbers for C#? What came out when? Why can't I find any answers about C# 3.5 ?
12 Answers
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How do I get indices of N maximum values in a NumPy array?
...
365
The simplest I've been able to come up with is:
In [1]: import numpy as np
In [2]: arr = np....
WinDBG用法详解 PDF - 文档下载 - 清泛网 - 专注C/C++及内核技术
...DBG具有非常大的灵活性和可扩展性，用来满足各种
第30章WinDBG用法详解...................................................................................................1
30.1工作空间......................................................................................
Removing projects in Sublime Text 2 and 3
How do you remove a project from Sublime Text 2 and 3's project windows ( Ctrl + Alt + P ) ?
5 Answers
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Output data from all columns in a dataframe in pandas [duplicate]
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Erel Segal-Halevi
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answered Nov 5 '12 at 18:13
YarivYariv
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[since C++11] std::array的使用 - C/C++ - 清泛网 - 专注C/C++及内核技术
...类、没有虚函数.
因此不支持这样的构造方法：array<int, 3> a({1, 2, 4});
初始化array最常用的方法是使用赋值运算符和初始化列表：
array<int, 3> a = {1, 2, 3};
array<int, 100> b = {1, 2, 3}; // a[0] ~ a[2] = 1, 2, 3; a[3] ~ a[99] = 0, 0, 0 ... 0;
array<in...
Single Line Nested For Loops
...f the list comprehension (here (x,y)):
>>> [(x, y) for x in [1,2,3] for y in [3,1,4] if x != y]
[(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)]
It's exactly the same as this nested for loop (and, as the tutorial says, note how the order of for and if are the same).
>>>...
Reorder levels of a factor without changing order of values
...f <- data.frame(f = 1:4, g = letters[1:4])
df
# f g
# 1 1 a
# 2 2 b
# 3 3 c
# 4 4 d
levels(df$g)
# [1] "a" "b" "c" "d"
df$g <- factor(df$g, levels = letters[4:1])
# levels(df$g)
# [1] "d" "c" "b" "a"
df
# f g
# 1 1 a
# 2 2 b
# 3 3 c
# 4 4 d
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