[course]03 —— Python循环
EX32
the_count = [1, 2, 3, 4, 5]
fruits = ['apples', 'oranges', 'pears', 'apricots']
change = [1, 'pennies', 2, 'dimes', 3, 'quarters']
# this first kind of for-loop goes through a list
p = 0
# 迭代 iterator
for number in the_count:
print("This is count %d" % number)
print(number)
print(p)
# same as above
for fruit in fruits:
print("A fruit of type: %s" % fruit)
# also we can go through mixed lists too
# notice we have to use %r since we don't know what's in it
for i in change:
print("I got %r" % i)
# we can also build lists, first start with an empty one
elements = []
# then use the range function to do 0 to 5 counts
for i in range(0, 6):
print("Adding %d to the list." % i)
# append is a function that lists understand
elements.append(i)
print(elements)
# now we can print them out too
for i in elements:
print("Element was: %d" % i)
index = 0
while index < len(elements):
print(elements[index])
index += 1EX33
i = 0
numbers = []
while i < 6:
print("At the top i is %d" % i)
numbers.append(i)
i = i + 2
if i == 4:
continue
print("Numbers now: ", numbers)
print("At the bottom i is %d" % i)
print("The numbers: ")
for num in numbers:
print(num)1. for循环和range
range方法的使用，第一个参数指定开始，第二个参数指定到哪里结束(<)
# A for loop repeats an action a specific number of times
# based on the provided range
def sumFromMToN(m, n):
total = 0
# note that range(x, y) includes x but excludes y
for x in range(m, n+1):
total += x
return total
print(sumFromMToN(5, 10) == 5+6+7+8+9+10)不使用range的计算方式
def sumFromMToN(m, n):
return sum(range(m, n+1))
print(sumFromMToN(5, 10) == 5+6+7+8+9+10)
# And we can even do this with a closed-form formula,
# which is the fastest way, but which doesn't really
# help us demonstrate loops. :-)
def sumToN(n):
# helper function
return n*(n+1)//2
def sumFromMToN_byFormula(m, n):
return (sumToN(n) - sumToN(m-1))
print(sumFromMToN_byFormula(5, 10) == 5+6+7+8+9+10)如果range只写一个参数则默认从0开始
def sumToN(n):
total = 0
# range defaults the starting number to 0
for x in range(n+1):
total += x
return total
print(sumToN(5) == 0+1+2+3+4+5)Copy Visualize Run
range的第三个参数代表步长(step)
def sumEveryKthFromMToN(m, n, k):
total = 0
# the third parameter becomes a step
for x in range(m, n+1, k):
total += x
return total
print(sumEveryKthFromMToN(5, 20, 7) == (5 + 12 + 19))计算m到n之间的奇数之和
# We can also change the step by changing the inside of the loop
def sumOfOddsFromMToN(m, n):
total = 0
for x in range(m, n+1):
if (x % 2 == 1):
total += x
return total
print(sumOfOddsFromMToN(4, 10) == sumOfOddsFromMToN(5,9) == (5+7+9))使用步长来计算
# Here we will range in reverse
# (not wise in this case, but instructional)
def sumOfOddsFromMToN(m, n):
total = 0
for x in range(n, m-1, -1):
if (x % 2 == 1):
total += x
return total
print(sumOfOddsFromMToN(4, 10) == sumOfOddsFromMToN(5,9) == (5+7+9))2. 循环嵌套
循环可以嵌套使用，在循环嵌套的时候更要注意两次循环的次数
# We can put loops inside of loops to repeat actions at multiple levels
# This prints the coordinates
def printCoordinates(xMax, yMax):
for x in range(xMax+1):
for y in range(yMax+1):
print("(", x, ",", y, ") ", end="")
print()
printCoordinates(4, 5)使用循环来写*
def printStarRectangle(n):
# print an nxn rectangle of asterisks
for row in range(n):
for col in range(n):
print("*", end="")
print()
printStarRectangle(5)# What would this do? Be careful and be precise!
def printMysteryStarShape(n):
for row in range(n):
print(row, end=" ")
for col in range(row):
print("*", end=" ")
print()
printMysteryStarShape(5)3. while循环
# use while loops when there is an indeterminate number of iterations
def leftmostDigit(n):
n = abs(n)
while (n >= 10):
n = n//10
return n
print(leftmostDigit(72658489290098) == 7)Example: nth positive integer with some property
# eg: find the nth number that is a multiple of either 4 or 7
def isMultipleOf4or7(x):
return ((x % 4) == 0) or ((x % 7) == 0)
def nthMultipleOf4or7(n):
found = 0
guess = -1
while (found <= n):
guess += 1
if (isMultipleOf4or7(guess)):
found += 1
return guess
print("Multiples of 4 or 7: ", end="")
for n in range(15):
print(nthMultipleOf4or7(n), end=" ")
print()Misuse: While loop over a fixed range
# sum numbers from 1 to 10
# note: this works, but you should not use "while" here.
# instead, do this with "for" (once you know how)
def sumToN(n):
# note: even though this works, it is bad style.
# You should do this with a "for" loop, not a "while" loop.
total = 0
counter = 1
while (counter <= n):
total += counter
counter += 1
return total
print(sumToN(5) == 1+2+3+4+5)4. break 和 continue
break： 结束当前循环 continue： 跳出当次循环，继续执行下一次循环
# continue, break, and pass are three keywords used in loops
# in order to change the program flow
for n in range(200):
if (n % 3 == 0):
continue # skips rest of this pass
elif (n == 8):
break # skips rest of entire loop
else:
pass # does nothing! pass is a placeholder, not needed here
print(n, end=" ")
print()Copy Visualize Run
while true循环的跳出
# Note- this is advanced content, as it uses strings. It's okay
# to not fully understand the content below.
def readUntilDone():
linesEntered = 0
while (True):
response = input("Enter a string (or 'done' to quit): ")
if (response == "done"):
break
print(" You entered: ", response)
linesEntered += 1
print("Bye!")
return linesEntered
linesEntered = readUntilDone()
print("You entered", linesEntered, "lines (not counting 'done').")5. 素数
使用循环来求素数
# Note: there are faster/better ways. We're just going for clarity and simplicity here.
def isPrime(n):
if (n < 2):
return False
for factor in range(2,n):
if (n % factor == 0):
return False
return True
# And take it for a spin
for n in range(100):
if isPrime(n):
print(n, end=" ")
print()更快的查询素数的方式
# Note: this is still not the fastest way, but it's a nice improvement.
def fasterIsPrime(n):
if (n < 2):
return False
if (n == 2):
return True
if (n % 2 == 0):
return False
maxFactor = round(n**0.5)
for factor in range(3,maxFactor+1,2):
if (n % factor == 0):
return False
return True
# And try out this version:
for n in range(100):
if fasterIsPrime(n):
print(n, end=" ")
print()验证
def isPrime(n):
if (n < 2):
return False
for factor in range(2,n):
if (n % factor == 0):
return False
return True
def fasterIsPrime(n):
if (n < 2):
return False
if (n == 2):
return True
if (n % 2 == 0):
return False
maxFactor = round(n**0.5)
for factor in range(3,maxFactor+1,2):
if (n % factor == 0):
return False
return True
# Verify these are the same
for n in range(100):
assert(isPrime(n) == fasterIsPrime(n))
print("They seem to work the same!")
# Now let's see if we really sped things up
import time
bigPrime = 499 # Try 1010809, or 10101023, or 102030407
print("Timing isPrime(",bigPrime,")", end=" ")
time0 = time.time()
print(", returns ", isPrime(bigPrime), end=" ")
time1 = time.time()
print(", time = ",(time1-time0)*1000,"ms")
print("Timing fasterIsPrime(",bigPrime,")", end=" ")
time0 = time.time()
print(", returns ", fasterIsPrime(bigPrime), end=" ")
time1 = time.time()
print(", time = ",(time1-time0)*1000,"ms")6. 计算第n个素数
def isPrime(n):
if (n < 2):
return False
if (n == 2):
return True
if (n % 2 == 0):
return False
maxFactor = round(n**0.5)
for factor in range(3,maxFactor+1,2):
if (n % factor == 0):
return False
return True
# Adapt the "nth" pattern we used above in nthMultipleOf4or7()
def nthPrime(n):
found = 0
guess = 0
while (found <= n):
guess += 1
if (isPrime(guess)):
found += 1
return guess
# and let's see a list of the primes
for n in range(10):
print(n, nthPrime(n))
print("Done!")Last updated
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