planets = ['Mercury', 'Venus', 'Earth', 'Mars', 'Jupiter', 'Saturn', 'Uranus', 'Neptune']
planet_to_initial = {planet: planet[0] for planet in planets}
planet_to_initial

for loop over a dictionary will loop over its keys

for k in numbers:
print("{} = {}".format(k, numbers[k]))

We can access a collection of all the keys or all the values with dict.keys() and dict.values(), respectively.

' '.join(sorted(planet_to_initial.values()))

The very useful dict.items() method lets us iterate over the keys and values of a dictionary simultaneously. (In Python jargon, an item refers to a key, value pair)

for planet, initial in planet_to_initial.items():
print("{} begins with \"{}\"".format(planet.rjust(10), initial))
Mercury begins with "M"
Venus begins with "V"
Earth begins with "E"
Mars begins with "M"
Jupiter begins with "J"
Saturn begins with "S"
Uranus begins with "U"
Neptune begins with "N"

--

--

datestr = '1956-01-31'
year, month, day = datestr.split('-')
'/'.join([month, day, year])

Farklı bitane

claim = "Pluto is a planet!"
words = claim.split() # default olarak boşluklardan ayırır

# Yes, we can put unicode characters right in our string literals :)

' 👏 '.join([word.upper() for word in words])

print içinde genelde kullandığımız yöntem

"{}, you'll always be the {}th planet to me.".format(planet, position)

üsttekinin ileri seviye kullanımı

pluto_mass = 1.303 * 10**22
earth_mass = 5.9722 * 10**24
population = 52910390
# 2 decimal points 3 decimal points, format as percent separate with commas
"{} weighs about {:.2} kilograms ({:.3%} of Earth's mass). It is home to {:,} Plutonians.".format(
planet, pluto_mass, pluto_mass / earth_mass, population,
)

# Referring to format() arguments by index, starting from 0
s = """Pluto's a {0}.
No, it's a {1}.
{0}!
{1}!""".format('planet', 'dwarf planet')
print(s)

--

--

[
planet.upper() + '!'
for planet in planets
if len(planet) < 6
]

tek satır hali

def count_negatives(nums):
return len([num for num in nums if num < 0])
print(count_negatives([5, -1, -2, 0, 3]))

boolean kullanımı

def count_negatives(nums):
# Reminder: in the "booleans and conditionals" exercises, we learned about a quirk of
# Python where it calculates something like True + True + False + True to be equal to 3.
return sum([num < 0 for num in nums])

--

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