CONTINUAL INTEGRATION

Problem scenario
List operations such as .sort() will change previous copies of the list.

String operations such as .replace("old_pattern", "new_pattern", count), will not change the variable or its copies; these operations will return a string after the replace operation has run on the original string.

Why is this behavior happening?

Solution
In-place changes and mutability are factors. Strings are generally immutable whereas copies of lists are usually shallow.

program1.py shows how lists are mutable with inplace changes happening to a copy; program2.py shows how strings are immutable and have no inplace changes to them when the .replace method is used.

# Filename: program1.py

one_list = ['dog', 'cat', 'horse', 'cow']
print(one_list)
print("Above is one_list before modifications have happened.")
copy_of_list = one_list
one_list.append('hamster')
one_list.sort()
print("The one_list was modified earlier in the execution of this program.")
print(one_list)
print("The one_list is above and the copy_list is below")
print(copy_of_list)
print("The .append and .sort methods change copies of the string in addition to the string itself.")
print("You may want to use deep copies instead of shallow copies if you don't want copies to be modified with the original.")

---

# Filename: program2.py:

first_string = "aaaaaaaaaaaaaaaaaabcdefgh"
second_string = first_string.replace('a', 'z', -1)
print("The replace method was invoked earlier in the execution of this program.")
print(" ")
print(first_string)
print("first_string is above; second_string is below")
print(second_string)

print("Now you can see how the .replace method only emphemerally returns a string after operating on it.")
print("The .replace method does not change the string it operates on.  The operations can be saved to a new variable if you want to use .replace.")

The replace method of a string does not modify the variable it operates on. It returns the modified string. With list manipulations, the copy of the list can be changed (unless you copy like this new_list = old_list[:]).

Problem scenario
You are running a Python script. You keep getting "SyntaxError: invalid syntax" and a line number. You can find nothing wrong with the syntax on the specified line number. What do you do?

Possible solution #1
Look at the line with code, as opposed to blank line(s), above it. Is there a missing quote, brace, bracket or parentheses?

Possible solution #2
Are you trying to use an "if" statement on one line? Make sure you have an "else" statement. Without an "else" clause, you can get "SyntaxError: invalid syntax". Here are four lines of code that are minimally correctly written:

f = 5
a = "unique"
x = f
print("scissors" if a == "unique" else "rocks")

# You could try to eliminate the 'else "rocks"' portion and run the program to see the error.

Problem scenario
You have a list in Python that you copy into another variable. You change the original list and the copied version changes. Why does the copied version get copied and what can you do to overcome this?

Solution
There is something called a "deep" copy and a "shallow" copy. This program will illustrate both the problem and the solution:

list_a = [3, 6, 7]
list_b = list_a
list_c = list_a[:]
list_a.append(4)
print(list_b)
print("list_b, a shallow copy, is above; list_a is below")
print(list_a)
print("a different copy of list_a is below")
print(list_c)

For more information, read the following links: StackOverflow.com and geeksforgeeks.com.