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Convert Array to Object
... It will return the target object.
Object.assign({}, ['a','b','c']); // {0:"a", 1:"b", 2:"c"}
The own length property of the array is not copied because it isn't enumerable.
Also, you can use ES6 spread syntax to achieve the same result:
{ ...['a', 'b', 'c'] }
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Python dict how to create key or append an element to key?
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edited Sep 30 '14 at 22:31
answered Oct 16 '12 at 0:43
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How do you clone a BufferedImage
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answered Aug 18 '10 at 16:26
KlarkKlark
7,59233 gold badges3232 silver badges5757 bronze badges
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Is calculating an MD5 hash less CPU intensive than SHA family functions?
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Yes, MD5 is somewhat less CPU-intensive. On my Intel x86 (Core2 Quad Q6600, 2.4 GHz, using one core), I get this in 32-bit mode:
MD5 411
SHA-1 218
SHA-256 118
SHA-512 46
and this in 64-bit mode:
MD5 407
SHA-1 312
SHA-256 148
SHA-512 189
Figures are in megabytes ...
How to determine function name from inside a function
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You can use ${FUNCNAME[0]} in bash to get the function name.
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Calling clojure from java
...intln (str "(binomial 5 3): " (binomial 5 3)))
(println (str "(binomial 10042 111): " (binomial 10042 111)))
)
If you run it, you should see something like:
(binomial 5 3): 10
(binomial 10042 111): 49068389575068144946633777...
And here's a Java program that calls the -binomial function in th...
ActiveRecord: size vs count
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edited Jul 10 '12 at 22:15
Jo Liss
22.5k1414 gold badges101101 silver badges150150 bronze badges
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Add missing dates to pandas dataframe
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270
You could use Series.reindex:
import pandas as pd
idx = pd.date_range('09-01-2013', '09-30-201...
List of installed gems?
... Gem::Requirement.default)
specs = Gem.source_index.search(dep)
puts specs[0..5].map{ |s| "#{s.name} #{s.version}" }
# >> Platform 0.4.0
# >> abstract 1.0.0
# >> actionmailer 3.0.5
# >> actionpack 3.0.5
# >> activemodel 3.0.5
# >> activerecord 3.0.5
Here's an ...
How do you add an array to another array in Ruby and not end up with a multi-dimensional result?
...note the asterisk, and that a2 is the receiver
or splice:
a1[a1.length, 0] = a2
a1[a1.length..0] = a2
a1.insert(a1.length, *a2)
or append and flatten:
(a1 << a2).flatten! # a call to #flatten instead would return a new array
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