After a whirlwind introduction to Python you've made it to the first checkpoint. This isn't about us checking up on you, but rather about you having a chance to check your own understanding of the foundational concepts covered so far. It is really easy to click through the coding cells, become a bit flummoxed, look at the answer quickly 'just to check', and then come away with the impression that you've understood what you are doing: "Yeah, yeah, I made a little mistake there but it's fine because I was close enough..."
We know that's the case because we've done it before you.
The challenge with coding is that, as the code becomes more complex, so do the issues. And if you haven't learned to be careful with the basics and to fix them yourself then fixing the more subtle 'mistakes' will be that much harder. And you don't want this to be any harder than it alreay is!
Everyone makes mistakes. We do it all the time because, in programming, it is very easy to make errors. Computers are stupid and do exactly what we tell them, so even tiny mistakes (aVariable vs avariable, for example) can cause a piece of code or a program to fail to run. Errors in a program are often called bugs.
Fun fact: the origin of the term bug was an actual bug (a moth) caught up in the mechanical relay of an early computer.
Bugs today are rarely of the insectile variety, and we now distinguish between three general types of error:
Here's a nice little video on this topic...
When you first learn to code, syntax errors are the most common type of error because you are still learning the 'rules' of the language: indentation, the difference between = and ==, the need for a colon (:) on the end of some types of code... Python is fairly good at telling you what has caused a syntax error when it's trivial and happens on a single line of code, but it will get quite a bit harder as your code becomes more complex!
The other types of errors are ones that you learn to spot over time through experience and developing skills. Learning to spot logic errors is one of those wonderful transferrable skills that you can also use in everyday life: by learning to think through challenges at a higher level and then address them iteratively you are likely to become both a more careful thinker (i.e. how should I go about solving this issue?), and a more flexible and resilient thinker (i.e. ok, so that didn't work out as I expected, what should do I do now?).
“A good example of a logic error might be, 'take a drink from your water bottle, put it in your backpack, walk to the library, and then put the top back on the bottle.'” Severance (2014, p.14)
Many sites don't even discuss the concept of a semantic error or treat it as the equivalent of a logic error, but we think it's useful to set a bar beyond 'basic' logic errors: these arise from interactions between parts of a program, or between a programming language and the underlying computer/operating system. So the code can (and usually does) run successfully but the output is a long way from what you expected, or the failures that you're seeing don't make sense. We'd suggest that spotting semantic errors is a sign that you are well on your way to thinking computationally. In short, you understand why your code is triggering strange and unexpected behaviour even though it looks like it should do exactly what you intended.
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x = 12
y = 10
z = x + y
x = x + y
y = z + y
Once you have calculated what you think x, y and z are, add print statements to the code above and run it to check. Make sure you understand the results you find.
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x = 1 + 2 * 3 / 4
y = (1 + 2) * (3 / 4)
z = 1 + 2 * (3 / 4)
print(x)
print(y)
print(z)
Once you have calculated what you think x, y and z are, add print statements to the code above and run it to check.
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pi = 3.141592653589793
r = 12 / 2
vol = (4/3) * pi * (r**3)
print(vol)
Given a sphere of volume 14,137cm3 calculate its radius to the closest whole number (this will involve rearranging the formula and you may need to good how to find a cube root with python):
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vol=14137
r3 = vol / ((4/3) * pi)
r = r3**(1/3)
print(round(r))
This is not a loop question, we just want you to change the value of hours so that for each of the cases below (hours is set to 10, 2, 0 in turn) you get the right output.
Using the code below as a starting point, add two lines of code so that your output is as follows:
Hours are greater than zero when hours=10Hours are greater than zero and Hours are less than 10 when hours=2 Hours are less than 10 when hours=0
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hours=10
if hours > 0:
print(" ")
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hours=2
if hours > 0:
print("Hours are greater than zero")
if hours < 10:
print("Hours are less than 10")
In the code cell below, enter code to execute the flow chart shown in the image below. You will need to assign values to a and b on lines 1 and 2 before using the flow chart to complete the code such that it produces the following:
a = 2 and b = 2 four lines of output will be writtena = 1 and b = 2 one line of output will be written
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a = 1
b = 2
x = a
if x == b:
print("inside 1")
print("inside 2")
print("inside 3")
print("after condition")
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x = 1
if x > 0 and x < 5:
print("Joe")
if x > 0 or x < 5:
print("Aled")
if not(x > 0):
print("Sarah")
Assume that the code executes for a given value of x and answer the following questions. Only test the code AFTER you have worked out the answers and typed them below.
x = 5? x be when the names Joe and Aled are printed? x = -1?Answers Edit this cell!
To edit this cell just double click on it (right here ). When you're finished, click the run cell button in the toolbar just below the menu near the top of the window.
1.
2.
3.
4.
Aled
1, 2, 3, 4
Aled, Sarah
No
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for x in range(0,9):
# ... do something ...
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for x in range(0,9):
print(x)
if(x%2 == 0):
print("x is even")
else:
print("x is odd")
print(x)
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if x != y: #line 1
print("x is not equal to y")
if(x > y): #line 4
print("x is greater than y")
else: #line 7
print("x is less than y")
else:
print("insert conclusion here")
Note how the indentation makes it easier to work out which ‘level’ the code is operating on. In the code above, lines 4 and 7 are at the same indentation meaning that both will be skipped if the initial condition (on line 1) is False.
To check you understand how the code above works:
insert conclusion here in the code above to a string that helps to explain the condition of x and yx = 2 and y = 3, type below what line(s) will be output Output for x = 2 and y = 3 is:
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x = 2
y = 3
if x != y: #line 1
print("x is not equal to y")
if(x > y): #line 4
print("x is greater than y")
else: #line 7
print("x is less than y")
else:
print("x is equal to y")
In the code cell below, enter code that reports whether person A is older than person B or not. Whole numbers should be used as ages in years and these ages should be specified on the first two lines of code. Either “Person A is older than Person B” or “Person A is not older than Person B” should be reported to the user.
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ageA = 20
ageB = 19
if ageA > ageB:
print("Person A is older than Person B")
else:
print("Person A is not older than Person B")
Copy your final code from the previous exercise into the code block below. Now build on this code to check if Person A is older than Person B, but also whether they are the same age. One of following three responses should be reported to the user:
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ageA = 20
ageB = 20
if ageA > ageB:
print("Person A is older than Person B")
elif ageA < ageB:
print("Person B is older than Person A")
else:
print("Person A and Person B are the same age")
Copy your final code from the previous exercise into the code block below. Build on this code with a nested execution structure to report if Person A is much younger, a little younger, the same age, a little older or much older than Person B. Write your code with a nested structure so that first it evaluates whether one person is older than the other, and then, if so, whether there is a large difference in age or a small difference in age. You should create a variable to specify what the number of years is that indicates a ‘large difference’ in age. One of the following five responses should be reported to the user:
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ageA = 25
ageB = 30
largeD = 20
diff = ageA - ageB
if diff <= (largeD * -1):
print("Person A is much younger than Person B")
elif diff < 0 and diff > (largeD * -1):
print("Person A is a little younger than Person B")
elif diff == 0:
print("Person A is the same age as Person B")
elif diff >= (largeD * 1):
print("Person A is much older than Person B")
elif diff > 0 and diff < (largeD * 1):
print("Person A is a little older than Person B")
Python provides tools to 'handle exceptions' (i.e. manage errors). For example you can use combinations of try, except, else and finally keywords.
Read the link above (and maybe Google yourself for some other examples) and use the information gained to handle an exception in which the user an age as a string rather than a integer (e.g. print "Invalid Input" if diff cannot be calculated):
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ageA = "five"
ageB = 30
largeD = 20
try:
diff = ageA - ageB
except:
print("Invalid Input")
else:
if diff <= (largeD * -1):
print("Person A is much younger than Person B")
elif diff < 0 and diff > (largeD * -1):
print("Person A is a little younger than Person B")
elif diff == 0:
print("Person A is the same age as Person B")
elif diff >= (largeD * 1):
print("Person A is much older than Person B")
elif diff > 0 and diff < (largeD * 1):
print("Person A is a little older than Person B")
#read more at https://docs.python.org/3/tutorial/errors.html#handling-exceptions
Another way to trap errors is by 'raising exceptions'. This requires using the raise keyword.
Read the link above (and maybe Google yourself for some other examples) and use the information gained to raise an exception if the user provide a negative age (outputing "Invalid Age" in the error mmessage):
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ageA = -1
ageB = 30
largeD = 20
diff = ageA - ageB
if ageA < 0 or ageB < 0:
raise Exception("Invalid Age")
else:
if diff <= (largeD * -1):
print("Person A is much younger than Person B")
elif diff < 0 and diff > (largeD * -1):
print("Person A is a little younger than Person B")
elif diff == 0:
print("Person A is the same age as Person B")
elif diff >= (largeD * 1):
print("Person A is much older than Person B")
elif diff > 0 and diff < (largeD * 1):
print("Person A is a little older than Person B")
#read more at https://docs.python.org/3/tutorial/errors.html#raising-exceptions
The following individuals have contributed to these teaching materials:
The content and structure of this teaching project itself is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 license, and the contributing source code is licensed under The MIT License.
Supported by the Royal Geographical Society (with the Institute of British Geographers) with a Ray Y Gildea Jr Award.
This notebook may depend on the following libraries: None