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How to convert these 3 values together into age?

By : user2175770
Date : October 14 2020, 08:10 PM
wish help you to fix your issue We have 3 values separately birthday birthmonth and birthyear since in the front end each input is separate. In the back-end where I use node js we want to get age from these 3 details. we found such a usage of moment js in here: moment().diff(moment('20170507', 'YYYYMMDD'), 'years') But my code is a bit different for using such a result. , Try
code :
const age = moment().diff(moment([year, month - 1, day]), 'years');

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Using PIL and NumPy to convert an image to Lab array, modify the values and then convert back

By : Michaela Biggie Germ
Date : March 29 2020, 07:55 AM
Hope this helps Without having tried it, scaling errors are common in converting colors:
RGB is bytes 0 .. 255, e.g. yellow [255,255,0], whereas rgb2xyz() etc. work on triples of floats, yellow [1.,1.,0].
code :
    # unpack image array, 10 x 5 x 3 -> r g b --
img = np.arange( 10*5*3 ).reshape(( 10,5,3 ))
print "img.shape:", img.shape
r,g,b = img.transpose( 2,0,1 )  # 3 10 5
print "r.shape:", r.shape

    # pack 10 x 5 r g b -> 10 x 5 x 3 again --
rgb = np.array(( r, g, b )).transpose( 1,2,0 )  # 10 5 3 again
print "rgb.shape:", rgb.shape
assert (rgb == img).all()

    # rgb 0 .. 255 <-> float 0 .. 1 --
imgfloat = img.astype(np.float32) / 255.
img8 = (imgfloat * 255).round().astype(np.uint8)  
assert (img == img8).all()

Using cast or convert to convert negative int values to datetime in SQL Server

By : Daniel 'RTRD'
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further -53690 would be 1753-01-01. -53691 takes you out of the range of the datetime data type (which doesn't support dates before 1753). Casting -53690 works as your other examples do.

Why does datatype convert < long oldTimestamp = Convert.ToInt64(values) > return error ? unity c#

By : Kenny Blanckaert
Date : March 29 2020, 07:55 AM
this one helps. The error is telling you that TotalSeconds is a double not long. Simply replace long timestamp with double timestamp.
code :
double timestamp = (DateTime.UtcNow - new DateTime(1970, 1, 1, 0, 0, 0, 0)).TotalSeconds;
long timestamp = (long) (DateTime.UtcNow - new DateTime(1970, 1, 1, 0, 0, 0, 0)).TotalSeconds;
long timestamp = (DateTime.UtcNow - new DateTime(1970, 1, 1, 0, 0, 0, 0)).Ticks / TimeSpan.TicksPerSecond;

Ruby : Convert XML to JSON putting the values of Name as keys and the values of Contents as values

By : Sri naga mani kanta
Date : March 29 2020, 07:55 AM
wish of those help You can use XSLT to transform the XML and then use nokogiri to transform it:
The XSLT I used:
code :
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="resources">
    "resources": {
        <xsl:for-each select="string">
            "<xsl:value-of select="./@name"/>":
            "<xsl:value-of select="."/>"
                <xsl:when test="position() != last()">,</xsl:when>
require 'nokogiri'
document = Nokogiri::XML(File.read('input.xml'))
template = Nokogiri::XSLT(File.read('template.xslt'))
transformed_document = template.transform(document)

Long HEX Value Split to 2bytes Hex values or 4bytes Hex Values and convert it to decimal (float) values Using VB.Net

By : user1650455
Date : March 29 2020, 07:55 AM
I hope this helps you . I'm going to give this a shot as well, but do consider what I said on your newer question.
To read a certain amount of bytes and turn them into a number best is to use the BitConverter and its To*** methods.
code :
Dim StartIndex As Integer = 0
Dim EndIndex As Integer = bytes.Length - 1

'How many bytes to read: 2 or 4.
Dim BytesPerStep As Integer = If(C = 1, 2, 4)

'If the system uses little endianness we need to reverse the array and loop backwards.
If BitConverter.IsLittleEndian Then
    'Reverse the array.

    'Swap start and end index.
    StartIndex = EndIndex
    EndIndex = 0

    'Negate BytesPerStep: Go backwards.
    BytesPerStep = -BytesPerStep
End If

'Iterate the array <BytesPerStep> bytes at a time.
For i = StartIndex To EndIndex Step BytesPerStep
    If C = 1 Then
        'Read two bytes = Short (Int16).
        ListBox1.Items.Add(BitConverter.ToInt16(bytes, i))
        'Read four bytes = Single (Float)
        ListBox1.Items.Add(BitConverter.ToSingle(bytes, i))
    End If
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