如何以pythonic的方式做到这一点？

Question

Consider this Python snippet:

for a in range(10):

    if a == 7:
        pass
    if a == 8:
        pass
    if a == 9:
        pass
    else:
        print "yes"

How can it be written shorter?

#Like this or...
if a ?????[7,8,9]:
    pass

Answer 1

Use the in operator:

if a in (7,8,9):
    pass

Answer 2

To test if a falls within a range:

if 7 <= a <= 9:
  pass

To test if a is in a given sequence:

if a in [3, 5, 42]:
  pass

Answer 3

for a in range(10):
    if a > 6:
        continue
    print('yes')

Answer 4

Based on your original code the direct "pythonic" replacement is:

if not a in [7, 8, 9]:
     print 'yes'

or

if a not in [7, 8, 9]:
     print 'yes'

The latter reads a little better, so I guess it's a bit more "pythonic".

Answer 5

if a in [7,8,9]

Answer 6

Depending on what you want to do, the map() function can also be interesting:

def _print(x):
    print 'yes'

map(_print, [a for a in range(10) if a not in (7,8,9)])

Answer 7

What about using lambda.

>>> f = lambda x: x not in (7, 8, 9) and print('yes')
>>> f(3)
yes
>>> f(7)
False

Answer 8

Since the question is tagged as beginner, I'm going to add some basic if-statement advice:

 if a == 7: pass if a == 8: pass if a == 9: ... else: ... 

are three independent if statements and the first two have no effect, the else refers only to

  if a == 9: 

so if a is 7 or 8, the program prints "yes". For future use of if-else statement like this, make sure to use elif:

if a == 7:
    seven()
elif a == 8:
    eight()
elif a == 9:
    nine()
else:
    print "yes"

or use just one if-statement if they call for the same action:

if a == 7 or a == 8 or a == 9:
    seven_eight_or_nine()
else:
    print "yes"