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What is the colon operator in Ruby?
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:foo is a symbol named "foo". Symbols have the distinct feature that any two symbols named the same will be identical:
"foo".equal? "foo" # false
:foo.equal? :foo # true
This makes comparing two symbols really fast (sin...
How to disallow temporaries
For a class Foo, is there a way to disallow constructing it without giving it a name?
10 Answers
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Looping over arrays, printing both index and value
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You would find the array keys with "${!foo[@]}" (reference), so:
for i in "${!foo[@]}"; do
printf "%s\t%s\n" "$i" "${foo[$i]}"
done
Which means that indices will be in $i while the elements themselves have to be accessed via ${foo[$i]}
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How to decorate a class?
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Example:
def substitute_init(self, id, *args, **kwargs):
pass
class FooMeta(type):
def __new__(cls, name, bases, attrs):
attrs['__init__'] = substitute_init
return super(FooMeta, cls).__new__(cls, name, bases, attrs)
class Foo(object):
__metaclass__ = FooMeta
d...
How to select rows from a DataFrame based on column values?
... example,
import pandas as pd
import numpy as np
df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(),
'B': 'one one two three two two one three'.split(),
'C': np.arange(8), 'D': np.arange(8) * 2})
print(df)
# A B C D
# 0 foo one ...
代码块超过1.2w编译apk报错问题 - App Inventor 2 中文网 - 清泛IT社区，为创新赋能！
...所有平台都是点编译就服务器错误。
只有code服务器，能通过，但是编译过程报错：
RequestTooLargeError 可能指的是事务任务列表的大小以及数据存储实体的大小，
不能超过 1 MB。您要添加的任务加上数据的总大小是多少？h...
Overloading Macro on Number of Arguments
I have two macros FOO2 and FOO3 :
8 Answers
8
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Using property() on classmethods
...method property, create the property on the metaclass.
>>> class foo(object):
... _var = 5
... class __metaclass__(type): # Python 2 syntax for metaclasses
... pass
... @classmethod
... def getvar(cls):
... return cls._var
... @classmethod
... def s...
Creating an instance of class
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/* 1 */ Foo* foo1 = new Foo ();
Creates an object of type Foo in dynamic memory. foo1 points to it. Normally, you wouldn't use raw pointers in C++, but rather a smart pointer. If Foo was a POD-type, this would perform value-initial...
Namespace and class with the same name?
...d up in this unfortunate situation: you are writing Blah.DLL and
importing Foo.DLL and Bar.DLL, which, unfortunately, both have a type
called Foo:
// Foo.DLL:
namespace Foo { public class Foo { } }
// Bar.DLL:
namespace Bar { public class Foo { } }
// Blah.DLL:
namespace Blah
{
using Foo; ...