Make dictionary with for loop in python

Question

I want to create a dictionary in python with 40 key/value pairs. They keys should be a1, a2 ... a10, b1, b2 ... b10, c1, c2 ... c10, d1, d2 ... d10. I have a for loop that takes two variables and modifies them so they change after each loop. It loops 40 times to create 40 different values for each of the two variables.

In the dictionary, the value for the key a1, should be the value of x and y in the first loop. The value for the key a2 should be the value of x and y in the second loop. This continues to the key d10 which should have the value of x and y in the 40th loop.

How can I do this? I think I can use a for loop, but since I struggle with understanding for loops, I don't know what I should do.

Here is my code:

x = 200
y = 202

for i in range(40):
    if y != 202 + 9 * 84:
        y += 84
    else:
        x += 100
        y = y - 9 * 84

I want a dictionary that looks like this:

dictionary = {'a1':(200, 202), 'a2':(# the value of x and y after looping a second time, here it will be 200, 286)}

Banana · Accepted Answer · 2020-06-19 11:33:27Z

1

letters = "abcd"
dictionary = {}

x = 200
y = 202

for i in range(40):
    key = letters[int(i / 10)] + str(i % 10 + 1)
    dictionary[key] = x, y

    if y != 202 + 9 * 84:
        y += 84
    else:
        x += 100
        y = y - 9 * 84

answered Jun 19, 2020 at 11:33

Banana

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5 Comments

Grass Over a year ago

Traceback (most recent call last): File "/path/path/path/file.py", line 13, in <module> key = letters[int(i / 10 - 1)] + i % 10 TypeError: must be str, not int

Banana Over a year ago

@greenGrass Please try the new version.

Grass Over a year ago

It seems like it worked very well. I'll check to see if it did exactly what I wanted.

Grass Over a year ago

I will also see a little bit on the other answer to see if I can learn something

Grass Over a year ago

It looks like this did exactly what I needed.

Chase · Accepted Answer · 2020-06-19 11:42:46Z

Here's an elegant solution that keeps track of both the letter and the number, and formats them into the key using fstring.

x = 200
y = 202

# Create the dict first
res = {}

# Assign the letter to begin with
key_letter = 'a'

for i in range(0, 40):
    if i != 0 and i % 10 == 0:
        # Increment the letter since 10th element reached
        key_letter = chr(ord(key_letter) + 1)
    # Format the letter with the number
    res[f'{key_letter}{i % 10 + 1}'] = (x, y)
    if y != 202 + 9 * 84:
        y += 84
    else:
        x += 100
        y = y - 9 * 84
print(res)

Output:

{'a1': (200, 202), 'a2': (200, 286), 'a3': (200, 370), 'a4': (200, 454), 'a5': (200, 538), 'a6': (200, 622), 'a7': (200, 706), 'a8': (200, 790), 'a9': (200, 874), 'a10': (200, 958), 'b1': (300, 202), 'b2': (300, 286), 'b3': (300, 370), 'b4': (300, 454), 'b5': (300, 538), 'b6': (300, 622), 'b7': (300, 706), 'b8': (300, 790), 'b9': (300, 874), 'b10': (300, 958), 'c1': (400, 202), 'c2': (400, 286), 'c3': (400, 370), 'c4': (400, 454), 'c5': (400, 538), 'c6': (400, 622), 'c7': (400, 706), 'c8': (400, 790), 'c9': (400, 874), 'c10': (400, 958), 'd1': (500, 202), 'd2': (500, 286), 'd3': (500, 370), 'd4': (500, 454), 'd5': (500, 538), 'd6': (500, 622), 'd7': (500, 706), 'd8': (500, 790), 'd9': (500, 874), 'd10': (500, 958)}

This way you don't have to maintain a separate string to keep track of all the letters.

Nils Ole · Accepted Answer · 2020-06-19 11:55:58Z

The following is solely based on that the 40 iterations come from a-d x 1-10, so I got rid of your for-loop.

x = 200
y = 202
your_dict = {}

for j in ["a","b","c","d"]:
    for k in range(1,11):
        key = j + str(k)
        your_dict[key] = (x,y)
        if y != 202 + 9 * 84:
            y += 84
        else:
            x += 100
            y = y - 9 * 84

Result:

{'a1': (200, 202), 'a2': (200, 286), 'a3': (200, 370), 'a4': (200, 454), 'a5': (200, 538), 'a6': (200, 622), 'a7': (200, 706), 'a8': (200, 790), 'a9': (200, 874), 'a10': (200, 958), 'b1': (300, 202), 'b2': (300, 286), 'b3': (300, 370), 'b4': (300, 454), 'b5': (300, 538), 'b6': (300, 622), 'b7': (300, 706), 'b8': (300, 790), 'b9': (300, 874), 'b10': (300, 958), 'c1': (400, 202), 'c2': (400, 286), 'c3': (400, 370), 'c4': (400, 454), 'c5': (400, 538), 'c6': (400, 622), 'c7': (400, 706), 'c8': (400, 790), 'c9': (400, 874), 'c10': (400, 958), 'd1': (500, 202), 'd2': (500, 286), 'd3': (500, 370), 'd4': (500, 454), 'd5': (500, 538), 'd6': (500, 622), 'd7': (500, 706), 'd8': (500, 790), 'd9': (500, 874), 'd10': (500, 958)}

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