[GO] Techniques often used in python short coding (Notepad)

I will update the techniques I learned in show coding recently started on the site codefights from time to time.

Short coding consideration

• As a basic basis, it is important to first check the algorithm etc. before starting coding to understand the optimum solution (I was particularly keenly aware when I learned the formula after solving the number root by pushing it)
• There is also a pattern in which the answer is a decisive game.
• For example, when Input is 1 <= n <= 10 and the amount of calculation is huge, short coding starts after making a list of answers locally.
• A world where the omission of one character is a life
• If you get too absorbed in it, you will get crazy.

Basic

<And <=

The comparison operator <= can basically be omitted and can be reduced by ** 1 character **

# long
a <= 3

# short (When a is an integer)
a < 4

~-and-~

Corresponds to -1 and +1 respectively However, since it has a higher priority than the multiplication operator, it can be reduced by ** 2 characters ** by omitting the parentheses.

# long
b * (a - 1) * 5

# short
b * ~-a * 5

Ternary operator

: (Colon) and the left side of the assignment can be omitted

# long
if a < b:
    c = 4
else:
    c = 2

# short
c = 4 if a < b else 2

repr literal

(Python2 only) Use \ \ to convert to string

# long
str(1600)

# short
`1600`

True, False constant

If you want to return a logical value exactly, use a logical operation without using True or False.

# long
b = False

# short
b = 0 > 1



# long
return True

# short
return 1 > 0

import as Also shorten the module to import

# long
import math
math.sqrt(10)

# short
import math as m
m.sqrt(10)

Bulk equivalence check

Connect multiple string comparisons and perform at once

#Both the first and last characters match
# long
a[0] == b[0] and a[-1] == b[-1]

# short
a[0] + a[-1] == b[0] + b[-1]


#a and b,c and d have the same length(Or unique even when combined)If
# long
a == b and c == d

# short
a + c == b + d

Functions etc.

If it is a challenge of codefights, the answer in the form of a function will be tested.

lambda If possible, you can reduce ** 4 characters ** by converting the function to a lambda expression.

# long
def Hoge(n):
    return n * 2

# short
Hoge = lambda n: n * 2

Argument name

Argument names can be one character because there are many cases where they are not explicitly called.

# long
def Hoge(arg):
    return arg ** arg

# short
def Hoge(a):
    return a ** a

Recursive function

Assign to a one-character variable when using recursion

# long
def LongLongHoge(a):
    if a < 3:
        return a
    return LongLongHoge(a - 2) * LongLongHoge(a - 1)

# short
def LongLongHoge(a):
    if a < 3:
        return a
    return t(a - 2) * t(a - 1)
t = LongLongHoge

Even shorter with lambda and simultaneous assignment

# short
LongLongHoge = t = lambda a: a if a < 3 else t(a - 2) * t(a - 1)

application

Conditioning in a list

# long
t = a % 3
if t == 0:
    a = a * 2
elif t == 1:
    a = 0
elif t == 2:
    a = t * 2

# short
a = [a * 2, 0, (a % 3) * 2][a % 3]

# short
t = a % 3
a = [a * 2, 0, t * 2][t]

Use conditional expressions well

(a <b) returns True and False, but can be used for operations as 1, 0, respectively. There are many patterns that can be omitted by using it

# long
c = 4 if a < b else 2

# short
c = 2 + 2 * (a < b)

Create a function that contains other methods as they are

# long
myhex = lambda n: '%X' % n

# short
myhex = '{:X}'.format
myhex = '%X'.__mod__