CODEWITHSUNDEEP
pythonstring
Rahul Agarwal
Rahul Agarwal

Reputation: 4100

Pass a string to RGB Colour format

I have a dict like this:

colour_dict={'Red': '0xFF, 0x00, 0x00','Green':'0x00, 0xFF, 0x00','White':'0xFF, 0xFF, 0xFF','Yellow':'0xff, 0xff, 0x00'}

I need to pass value of this dict to this code and it works right:

from pptx.dml.color import RGBColor
RGBColor(0xFF, 0x00, 0x00)

When I do below to achieve the same:

RGBColor(colour_dict["Red"])

It throws me an error as this is a string and this just want in straight number format.

I already tried converting this string to int format but nothing is working.

Overall Problem:

How to pass the value to RGBColor using dict?

Upvotes: 0

Views: 2812

Answers (4)

GPhilo
GPhilo

Reputation: 19123

colour_dict["Red"] is the string "0xFF, 0x00, 0x00", which of course can't be directly converted to an int.

You need to convert each piece in the string and pass those values to the RGBColor constructor:

color = RGBColor(*map(lambda v : int(v, 16), colour_dict["Red"].split(',')))

NOTE: There's a lot going on in that line, so I'll break it down and explain:

  • colour_dict["Red"].split(',') extracts the color values substrings: ['0xFF', '0x00', '0x00']

  • map() applies the function passed as first argument to each value in the iterable passed as second argument.

  • lambda v : int(v, 16) converts v from hex string to int

  • * before map mans "unpack the values in the iterable that follows". it's a convenient way to pass a list with the RGB values instead of calling RGBColor(R_val, G_val, B_val).

Python 2 version:

Since python 2 doesn't support the asterisk operator, you need to split the statement in two:

r,g,b = map(lambda v : int(v, 16), colour_dict["Red"].split(','))
color = RGBColor(r,g,b)

Upvotes: 2

Playphysics
Playphysics

Reputation: 1

colour_dict["Red"] is a string, not three numbers

you can do like this:

a = list(colour_dict["Red"])
RGBColor(int(a[0]), int(a[2]), int(a[4]))

a:['0xFF', ',', '0x00', ',', '0x00'], has 5 characters in total

Upvotes: 0

tevemadar
tevemadar

Reputation: 13195

I find the map lambda answer a bit overkill, here is a less fancy variant:

color = RGBColor(*[int(x,16) for x in colour_dict["Red"].split(",")])

Upvotes: 0

Tianbo Ji
Tianbo Ji

Reputation: 125

Replace

RGBColor(colour_dict["Red"])

by

eval("RGBColor({})".format(colour_dict["Red"]))

Upvotes: 0

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