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How can I obtain the element-wise logical NOT of a pandas Series?
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274
To invert a boolean Series, use ~s:
In [7]: s = pd.Series([True, True, False, True])
In [8]: ~...
What are the correct version numbers for C#?
...ession trees, anonymous types, implicit typing (var), query expressions
C# 4.0 released with .NET 4 and VS2010 (April 2010). Major new features: late binding (dynamic), delegate and interface generic variance, more COM support, named arguments, tuple data type and optional parameters
C# 5.0 released...
How do I get indices of N maximum values in a NumPy array?
... up with is:
In [1]: import numpy as np
In [2]: arr = np.array([1, 3, 2, 4, 5])
In [3]: arr.argsort()[-3:][::-1]
Out[3]: array([4, 3, 1])
This involves a complete sort of the array. I wonder if numpy provides a built-in way to do a partial sort; so far I haven't been able to find one.
If this ...
Output data from all columns in a dataframe in pandas [duplicate]
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42
There is too much data to be displayed on the screen, therefore a summary is displayed instead....
[since C++11] std::array的使用 - C/C++ - 清泛网 - 专注C/C++及内核技术
...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<int, 3> c; ...
Removing projects in Sublime Text 2 and 3
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334
It would be nice if Sublime Text removed projects you've deleted from the recent projects list. ...
Rspec: “array.should == another_array” but without concern for order
... answered Jun 5 '10 at 3:08
x1a4x1a4
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Split delimited strings in a column and insert as new rows [duplicate]
... of doing it..
df <- read.table(textConnection("1|a,b,c\n2|a,c\n3|b,d\n4|e,f"), header = F, sep = "|", stringsAsFactors = F)
df
## V1 V2
## 1 1 a,b,c
## 2 2 a,c
## 3 3 b,d
## 4 4 e,f
s <- strsplit(df$V2, split = ",")
data.frame(V1 = rep(df$V1, sapply(s, length)), V2 = unlist(...
Reorder levels of a factor without changing order of values
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Use the levels argument of factor:
df <- 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 ...