Automate the Boring Stuff Chapter 6 Table Printer Almost Done

Here's an alternate method that perhaps you could apply to your own code. I first took tableData and sorted it out into a dictionary so it's easier to work with. After that I found the longest list in terms of characters. This allows us to know how far over the shorter lists should go. Finally, I printed out each lists adding spaces in front of the shorter ones based on the difference from the longest.

# orginal datatableData=[['apples', 'oranges', 'cherries', 'banana'],        ['Alice', 'Bob', 'Carol', 'David'],        ['dogs', 'cats', 'moose', 'goose']]# empty dictonary for sorting the datanewTable = {0:[], 1:[], 2:[], 3:[]}# iterate through each list in tableDatafor li in tableData:    for i in range(len(li)):        # put each item of tableData into newTable by index        newTable[i].append(li[i])# determine the longest list by number of total characters# for instance ['apples', 'Alice', 'dogs'] would be 15 characters# we will start with longest being zero at the startlongest = 0# iterate through newTable# for example the first key:value will be 0:['apples', 'Alice', 'dogs']# we only really care about the value (the list) in this casefor key, value in newTable.items():    # determine the total characters in each list    # so effectively len('applesAlicedogs') for the first list    length = len(''.join(value))    # if the length is the longest length so far,    # make that equal longest    if length > longest:        longest = length# we will loop through the newTable one last time# printing spaces infront of each list equal to the difference# between the length of the longest list and length of the current list# this way it's all nice and tidy to the rightfor key, value in newTable.items():    print(' ' * (longest - len(''.join(value))) + ' '.join(value))

This is how I did.

For the first part of the code I just used the hint they give to us.

In Chapter 4 / Practice Project / Character Picture Grid we've learned how to "rotate" and then print a list of lists. It was useful for the second part of my code.

#!/usr/bin/python3# you can think of x and y as coordinatestableData = [['apples', 'oranges', 'cherries', 'banana'],             ['Alice', 'Bob', 'Carol', 'David'],             ['dogs', 'cats', 'moose', 'goose']]def printTable(table):    # create a new list of 3 "0" values: one for each list in tableData    colWidths = [0] * len(table)    # search for the longest string in each list of tableData    # and put the numbers of characters in the new list    for y in range(len(table)):        for x in table[y]:            if colWidths[y] < len(x):                colWidths[y] = len(x)    # "rotate" and print the list of lists    for x in range(len(table[0])) :        for y in range(len(table)) :            print(table[y][x].rjust(colWidths[y]), end = ' ')        print()        x += 1printTable(tableData)

Here you go young padawan:

tableData=[['apples', 'oranges', 'cherries', 'banana'],    ['Alice', 'Bob', 'Carol', 'David'],    ['dogs', 'cats', 'moose', 'goose']]maxlen = 0for fruit,name,animal in zip(tableData[0], tableData[1], tableData[2]):    maxlen = max(len(fruit) + len (name) + len (animal), maxlen)for fruit,name,animal in zip(tableData[0], tableData[1], tableData[2]):    length = len(fruit) + len (name) + len (animal)     print ((' ' * (maxlen - length)) + fruit, name, animal)

Looping to determine maxlen is probably not optimal, copypasting was just the quickest thing that came to my mind.