This is @aimof from iRidge Inc. I have hosted a mob programming study session within iRidge (mainly for young people), so I will summarize it when I report it. This time, I used the general code (FizzBuzz) instead of the production code. This article is a rewrite of a document shared internally by IRIDGE.
Since we were free to go in and out, there were some fluctuations, but the number of people was about 4 to 5 and we carried out for 2 hours. There are three members who participated from the beginning to the end.
It was good to immediately notice any misunderstandings or mistakes, as everyone other than the driver would have to think with each other with their heads together. (Although there were some bugs that weren't fixed: see below).
I think the members were able to work together with a sense of unity.
Here is the theme prepared this time.
FizzBuzz with additional requirements each time it's completed
For numbers from 1 to 100
Draw one from the requirements prepared in advance and add the requirements.
This time, I decided to put a box and a pen and ask them to write their requirements freely. Various things from basic requirements to difficult ones were gathered.
From here, I will tell you the atmosphere while writing the actual code.
for i in range(1, 101):
if i % (3 * 5) == 0:
print("FizzBuzz")
elif i % 3 == 0:
print("Fizz")
elif i % 5 == 0:
print("Buzz")
else:
print(i)
No commentary required? It's an ordinary FizzBuzz.
I gave up.
I pulled it in one shot because it was included as a story. It is very painful because basic syntax such as if, def, for is restricted. Next quickly.
Later, I was told one solution in-house. (It is python2.7)
i = 1
while i <= 100:
print "Fizz" * (not i % 3) + "Buzz" * (not i % 5) or i
i = i + 1
def has_six_different_num(values):
char_set = set()
for value in values:
if type(value) is int:
for char in str(value):
char_set.add(char)
if 6 <= len(char_set):
return True
else:
return False
values = list()
for i in range(1, 101):
value = ""
if i % (3 * 5) == 0:
value = "FizzBuzz"
elif i % 3 == 0:
value = "Fizz"
elif i % 5 == 0:
value = "Buzz"
else:
value = i
values.append(value)
if len(values) > 10:
del values[0]
if has_six_different_num(values):
value = "ok"
values[-1] = "ok"
print(value)
It has become quite long. The code is already worth refactoring. If you were writing code that would make you say that it's really unpleasant to turn for twice ...
I have drawn the requirement as if I was told to correct the double for statement. Everyone laughed unintentionally at the good timing.
Here are the results.
def has_six_different_num(values):
return 7 <= len(set(list("".join(values))))
values = list()
for i in range(1, 30001):
value = ""
if i % (3 * 5) == 0:
print("FizzBuzz")
elif i % 3 == 0:
print("Fizz")
elif i % 5 == 0:
print("Buzz")
else:
value = i
values.append(str(value))
if has_six_different_num(values[-10:]):
value = "ok"
values[-1] = "ok" # <-This is a bug
if value:
print(value)
It was quite refreshing. Since it is difficult to speed up with 100 times, the number to be output has increased to 30,000. It has increased from 0.044 seconds to 0.028 seconds.
7 <= len(set(list("".join(values))))
I learned a lot from this part.
Returns true if 7 types of char including the empty string are used.
Actually, this code bug. I just wrote it in the comments. If you do not substitute an empty string here, the two characters of ok will be treated as a number judgment.
def has_six_different_num(values):
return 7 <= len(set(list("".join(values))))
values = list()
for i in range(30000, 0, -1):
value = ""
if i % (3 * 5) == 0:
print("FizzBuzz")
elif i % 3 == 0:
print("Fizz")
elif i % 5 == 0:
print("Buzz")
else:
value = i
values.append(str(value))
if has_six_different_num(values[-10:]):
value = "ok"
values[-1] = "ok"
if value:
print(value)
Just change the range. I haven't noticed the bug yet.
rm -Rf ~ if a Runtime Error occursIt's gone for my convenience. It's a company computer ... By the way, the Runtime Error never happened after this!
Never run this command. The following home directories are deleted.
As I got used to it, the speed at which the driver could guess from Mob's remarks became faster. It seems that the line disappears immediately after saying that I want to erase it.
I forgot to save the code for this requirement, so I will summarize it in the next section.
import json
num = int(input())
def has_six_different_num(values):
return 7 <= len(set(list("".join(values))))
values = list()
output_values = list()
for i in range(num, 0, -1):
value = ""
if i % 3 == 0:
value += "Fizz"
if i % 5 == 0:
value += "Buzz"
if i % 7 == 0:
value += "Foo"
if i % 11 == 0:
value += "Bar"
if not value:
value = i
if type(value) is str:
values.append("")
else:
values.append(str(value))
if has_six_different_num(values[-10:]):
value = "ok"
values[-1] = ""
output_values.append(value)
print(json.dumps(output_values))
This is the final result. The bug has also been fixed.
It was a study session (trial session) for mob programming, but to be honest, I enjoyed it more than I expected. This time, I wrote a code with a strong event nature, so I think that the feeling that everyone was crazy was stronger than usual.
However, we also received feedback that we would like to use the production code if there is a next one. It's fun to prepare the code that is in trouble at IRIDGE and use it as a material for study sessions. (The problem is that you can't publish the code on Qiita)
In addition to the original purpose, it was a big merit that I could not do anything appropriate because it led to my own thinking.
I think it was good this time that the requirements will increase. The reason is that it is developed small and refactoring is very important.
Personally, I would like to have an even more interesting study session within iRidge. If we have an interesting study session, we will publish it on Qiita again!
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