<jupyter_start><jupyter_text><jupyter_code>my_number = 2 + 2 * 8 / 5.0
print(my_number) # 5.2
my_remainder = 9 % 4
print(my_remainder) # 1
two_squared = 2 ** 2
print(two_squared)    # 4
two_cubed = 2 ** 3
print(two_cubed)      # 8
string_one = "Hello,"
string_two = " World!"
combined = string_one + string_two
print(combined) # Hello, World!

lst_one = [1,2,3]
lst_two = [4,5,6]
big_lst = lst_one + lst_two
print(big_lst) # [1, 2, 3, 4, 5, 6]
my_string = "Bueller"
repeated = my_string * 3
print(repeated) # BuellerBuellerBueller

my_list = [1, 2, 3]
repeated_list = my_list * 3
print(repeated_list) # [1, 2, 3, 1, 2, 3, 1, 2, 3]
a = object()
b = object()
a_list = [a] * 5
b_list = [b] * 5
combined = a_list + b_list
print(len(combined)) # 10
"""
1. Assign two different types to the variables "a" and "b".

2. Use basic operators to create a list that contains three instances of "a" and three instances of "b".
"""
# Modify the code below to meet the requirements outlined above
a = "a"
b = "5"

my_list = a*3 + b*3
print(my_list)
[i.split('\t', 1)[0] for i in my_list]
for element in my_list:
    my_list = element.split(',')
    print(my_list)<jupyter_output>['a']
['a']
['a']
['5']
['5']
['5']
{'fraction_non_alphanumeric': 0.1590909090909091, 'fraction_numerical': 0.045454545454545456, 'mean_word_length': 3.6704545454545454, 'pattern_counts': {'":': 0, '<': 4, '<?xml version=': 0, '>': 4, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 0, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
{'chain_length': '1', 'dir': 'jupyter-structured-clean-dedup', 'id': '479693', 'license': 'MIT', 'n_tokens_mistral': 575, 'n_tokens_neox': 533, 'n_words': 151, 'path': '_BasicOperatorsStringsLists.ipynb', 'repo_name': 'nealonleo9/CS-Week-1', 'size': '1232'}