'How to make the value that goes to 0, to make it to go back to same value in python?
I'm using Python PIL module to draw some lines and I use for loop to change the width of the line. But now I would like to do, when line width is to 0, I want to go and add value of lw. Basicly when width value go to zero, then I want it to go back to the 120.
This is my code...
from PIL import Image, ImageDraw
img = Image.new('RGB', (2000, 2000), (0, 0, 0))
draw = ImageDraw.Draw(img)
lw = 0
for y in range(-100, 2100, 100):
lw = lw + 10
draw.line((2000, y, 0, y), (255,0,0), 120-lw)
img
This is an image from this code...
I was thinking to use if statement but I don't know where to incorporate it. Any ideas?
Thanks in advance!
Solution 1:[1]
for y in range(-100, 2100, 100):
lw = lw + 10
draw.line((2000, y, 0, y), (255,0,0), 120-lw)
Nice attempt.
I was thinking to use
ifstatement but I don't know where to incorporate it. Any ideas?
In this case I would suggest to do this:
for y in range(-100, 2100, 100):
lw = lw + 10
if (lw >= 120): # When lw reaches 120...
lw = 0 # ...you set it to 0
draw.line((2000, y, 0, y), (255,0,0), 120-lw)
Anyway a more elegant solution (which doesn't involve if statements) would be this:
for y in range(-100, 2100, 100):
lw = lw + 10
draw.line((2000, y, 0, y), (255,0,0), 120-(lw%120))
The % operator gets the remainder of lw/120, so that when lw>=120 you won't have a value greater than 120 anyway.
EDIT
I want to go from 120 to 0, and from 0 to 120. In this way it goes like this (120, 110,..., 20, 10, 0, 120, 110,...), but I would like to go like this (120, 110,..., 20, 10, 0, 10, 20,...,
In this case you have to use an if statement:
incrementing = True
lw = 0
for y in range(-100, 2100, 100):
if (lw == 120):
incrementing = False
elif (lw == 0):
incrementing = True
if (incrementing):
lw = lw + 10
else:
lw = lw - 10
draw.line((2000, y, 0, y), (255,0,0), 120-lw)
My suggestion is to use a boolean variable to make the incrementing/decrementing easier.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
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| Solution 1 |

