Suppose the
while
loop has an
else
clause.
When the expression next to the word
while
becomes false and we leave the loop,
the
else
clause will be executed.
If we leave the loop for any other reason (e.g.,
break,
sys.exit),
the
else
clause will not be executed.
while
loop to compute the
Fibonacci numbers.
break
and
continue
statements, and
else
clauses on loops
The indented statements will be repeated forever. Indent each indented statement with exactly four spaces.
It was a dark and stormy night. Some Indians were sitting by a campfire. Then their chief rose and said, It was a dark and stormy night. Some Indians were sitting by a campfire. Then their chief rose and said, It was a dark and stormy night. Some Indians were sitting by a campfire. Then their chief rose and said,
To stop the script,
click on IDLE’s “Python 3.8.0 Shell”
window and pull down
Shell →
Interrupt Execution
Traceback (most recent call last): File "/Users/myname/python/junk.py", line 15, in <module> time.sleep(3) #Do nothing for 3 seconds. KeyboardInterrupt >>>
Each iteration of the loop needs only one call to the
print
function.
The
triple-quoted
string
contains three lines,
each ending with one newline character.
The
print
outputs one newline after it outputs the string.
while True:
print("""\
It was a dark and stormy night.
Some Indians were sitting by a campfire.
Then their chief rose and said,
""")
time.sleep(3) #Do nothing for 3 seconds.
paragraph = """\
It was a dark and stormy night.
Some Indians were sitting by a campfire.
Then their chief rose and said,
"""
while True:
print(paragraph)
time.sleep(3) #Do nothing for 3 seconds.
You can put the boolean expression after the keyword
while
or after the keyword
if.
""" russianRoulette.py """ import sys import random instructions = """\ Welcome to Russian Roulette! Press return to spin the cylinder and pull the trigger.""" print(instructions) print() _ = input("Go ahead: ") #underscore is name for dummy variable n = random.randrange(6) #an integer in the range 0 to 5 inclusive while n != 0: print("Click.") print() _ = input("Go ahead: ") n = random.randrange(6) print("BANG!") sys.exit(0)
Welcome to Russian Roulette! Press return to spin the cylinder and pull the trigger. Go ahead: click Go ahead: click Go ahead: click Go ahead: click Go ahead: BANG!
No need to write the yellow section twice.
"""
russianRoulette.py
"""
import sys
import random
#a string containing 2 lines
instructions = """\
Welcome to Russian Roulette!
Press return to spin the cylinder and pull the trigger."""
print(instructions)
while True:
print()
_ = input("Go ahead: ")
n = random.randrange(6)
if n == 0:
break #Abandon the loop.
print("Click.")
#The break statement sends us here.
print("BANG!")
sys.exit(0)
See
else
and
break.
The
mantissa
is a fraction that is always greater than or equal to .5
and less than 1.
Please type a number: 12e Sorry, "12e" is not a number. Try again. Please type a number: q23 Sorry, "q23" is not a number. Try again. Please type a number: 123 Thank you for typing in the float number 123.0. It is stored internally as the pair of integers 8,655,355,533,852,672 and 7. 123.0 = (8,655,355,533,852,672 / (2 ** 53)) * (2 ** 7) = 0.9609375 * 128 The fraction 8,655,355,533,852,672 / (2 ** 53) is the mantissa. The integer 7 is the exponent.
Please type a number: 6 Thank you for typing in the float number 6.0. It is stored internally as the pair of integers 6,755,399,441,055,744 and 3. 6.0 = (6,755,399,441,055,744 / (2 ** 53)) * (2 ** 3) = 0.75 * 8 The fraction 6,755,399,441,055,744 / (2 ** 53) is the mantissa. The integer 3 is the exponent.
bc -l 6755399441055744 / 2^53 .75000000000000000000 control-d
Double the number each time around the loop.
2 ** 0 = 1 2 ** 1 = 2 2 ** 2 = 4 2 ** 3 = 8 2 ** 4 = 16 2 ** 5 = 32 2 ** 6 = 64 2 ** 7 = 128 2 ** 8 = 256 2 ** 9 = 512 2 ** 10 = 1024 2 ** 11 = 2048 2 ** 12 = 4096 2 ** 13 = 8192 2 ** 14 = 16384 2 ** 15 = 32768 2 ** 16 = 65536 2 ** 17 = 131072 2 ** 18 = 262144 2 ** 19 = 524288 The log of 1,000,000 to the base 2 is 19.931568569324174
It would be satisfying to also print the number that goes over a million.
For this,
we need a
break
statement in an
if
statement.
import sys
i = 0
p = 1
while True:
print(f"2 ** {i:2} = {p:9,}") #One million is a seven-digit number, plus 2 commas.
if p >= 1_000_000: #Test in the middle of the loop.
break
i += 1
p *= 2
sys.exit(0)
2 ** 0 = 1 2 ** 1 = 2 2 ** 2 = 4 2 ** 3 = 8 2 ** 4 = 16 2 ** 5 = 32 2 ** 6 = 64 2 ** 7 = 128 2 ** 8 = 256 2 ** 9 = 512 2 ** 10 = 1,024 2 ** 11 = 2,048 2 ** 12 = 4,096 2 ** 13 = 8,192 2 ** 14 = 16,384 2 ** 15 = 32,768 2 ** 16 = 65,536 2 ** 17 = 131,072 2 ** 18 = 262,144 2 ** 19 = 524,288 2 ** 20 = 1,048,576
Halve the number each time around the loop.
6.103515625e-05
means
6.103515625 × 10–5
=
6.103515625 × .00001
=
.00006103515625
The
e
stands for
“exponent”.
1.0
0.5
0.25
0.125
0.0625
0.03125
0.015625
0.0078125
0.00390625
0.001953125
0.0009765625
0.00048828125
0.000244140625
0.0001220703125
6.103515625e-05
3.0517578125e-05
1.52587890625e-05
7.62939453125e-06
3.814697265625e-06
1.9073486328125e-06
9.5367431640625e-07
4.76837158203125e-07
2.384185791015625e-07
1.1920928955078125e-07
etc.
2.2250738585072014e-308
1.1125369292536007e-308
5.562684646268003e-309
2.781342323134e-309
1.390671161567e-309
6.953355807835e-310
3.4766779039175e-310
1.73833895195875e-310
8.691694759794e-311
4.345847379897e-311
2.1729236899484e-311
1.086461844974e-311
5.43230922487e-312
2.716154612436e-312
1.35807730622e-312
6.7903865311e-313
3.39519326554e-313
1.69759663277e-313
8.487983164e-314
4.243991582e-314
2.121995791e-314
1.0609978955e-314
5.304989477e-315
2.65249474e-315
1.32624737e-315
6.63123685e-316
3.3156184e-316
1.6578092e-316
8.289046e-317
4.144523e-317
2.0722615e-317
1.036131e-317
5.180654e-318
2.590327e-318
1.295163e-318
6.4758e-319
3.2379e-319
1.61895e-319
8.095e-320
4.0474e-320
2.0237e-320
1.012e-320
5.06e-321
2.53e-321
1.265e-321
6.3e-322
3.16e-322
1.6e-322
8e-323
4e-323
2e-323
1e-323
5e-324
0.0
Number the lines of output.
Multiply the number by 1.04 each time around the loop.
#How many years would it take for my principal to double,
#at 4% compounded annually?
import sys
import math
year = 0
p = 100000.00 #principal
while True:
print(f"{year:2} ${p:,.2f}") #Print money with 2 digits to the right of the decimal point.
if p >= 200000.00:
break
year += 1
p *= 1.04
print()
print(f"The log to the base 1.04 of 2 is {math.log(2, 1.04)}")
sys.exit(0)
0 $100,000.00 1 $104,000.00 2 $108,160.00 3 $112,486.40 4 $116,985.86 5 $121,665.29 6 $126,531.90 7 $131,593.18 8 $136,856.91 9 $142,331.18 10 $148,024.43 11 $153,945.41 12 $160,103.22 13 $166,507.35 14 $173,167.64 15 $180,094.35 16 $187,298.12 17 $194,790.05 18 $202,581.65 The log to the base 1.04 of 2 is 17.672987685129698
2 = 21/12 × 21/12 × 21/12 × 21/12 × 21/12 × 21/12 × 21/12 × 21/12 × 21/12 × 21/12 × 21/12 × 21/12
Exponentiation has higher precedence than division. We therefore need the parens to force the division to be executed before the exponentiation.
>>> 2 ** (1/12) 1.0594630943592953
A picture of two octaves:
0 100.00000 inches 1 105.94631 inches 2 112.24620 inches 3 118.92071 inches 4 125.99210 inches 5 133.48399 inches 6 141.42136 inches 7 149.83071 inches 8 158.74011 inches 9 168.17928 inches 10 178.17974 inches 11 188.77486 inches 12 200.00000 inches
See the
while
loop in the
Teaser.
Please type an integer: 60 2 2 3 5
Better to write the loop in lines 13,
14, 20 as a
for
loop:
for factor in range(2, n + 1): #loop from 2 to n inclusive
while n % factor == 0: #while factor is a factor of n
print(factor)
n //= factor