We also note that the number of spaces before the star in any row = ((middle row star count+1)/2 - number of stars in the row).įor the second half of the diamond we repeat the same process. We notice that the number of stars in each row is (row_number*2)-1. For drawing this we first find how many spaces we need to draw and then how many stars we need to draw. In the first step we draw the upper diamond up to the middle row. We draw the diamond in two separate steps. # number of space before successive row is 1,2,3. # Note that we have one row less to print # Now draw below the middle row (lower diamond) # Each successive row has odd number of stars # We use python for string multiplication Spaces_before_star = max_spaces_needed - row # Each row has decreasing number of spaces # Note that python range excludes the upper limit. # First let us draw till middle row (upper diamond) Max_spaces_needed = (diamond_size+1)//2 - 1 For first row it is middle_row_index - 1. # We also note that the number of spaces needed in the first and last row # Note the use of integer divison in python using // The python for ASCII diamond generation is given below, # Number of rows needed in the diamond drawn We will split the problem by first printing half the diamond up to the middle row and then printing the bottom portion of the diamond separately. For example, for a diamond size of 5, the middle row is the (5+1)/2 = 3rd row. Writing this in python requires us to logically think about the solution and then use python programming structures such as loops to generate the ASCII diamond shape.įor a diamond of size n, the middle row number is (n+1)/2. Now let us try to write a python program which will print a diamond given the odd number of rows required in it. The number of stars increase and spaces decrease starting from the top and then from the middle row onwards the number of stars decrease and spaces increase. Looking at the diamond we can see that it is composed of varying numbers of "stars" and spaces.
For example, you can draw diamonds with 3, 5, 7, 9, 11 etc. For example, the following shows a diamond shape as ASCII art consisting of 9 rows, *įrom the pattern it can be seen that the smallest diamond is of 3 rows and symmetric diamonds of this shape will have odd number of rows. So let’s have a look at that as well.A common python programming problem given for beginners is to draw a diamond shape using star character as an ASCII art. You must be wondering about the output of our script. If you want to change your destination path, you can always set it in the target path like the following: pw.image_to_ascii_art('Capture.PNG', 'D:/TECHBIT/ASCII.txt') The converted image will always gets stored within the same folder.
Print("Converting your given image to ASCII Art") You can always customise the message or even completely skip it. The message reads “ Converting your given image to ASCII Art”.
#Beginner ascii art install
Make sure to install it beforehand using the above mentioned command. Now import the required module, i.e., pywhatkit.In your editor, set up a new project and then create a file with the name index.py.Store the image within the same folder to ease out the location path. An image which you wish to convert into ASCII art.Just install it like any other module, type in the following command in the terminal: pywhatkit module: This module is responsible for converting any image to ASCII art in no time.
This way it’ll become easy to manage and locate. Make sure to make a new folder to store all the related files of this particular project.
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