Python의 조건, 루프 등에 대한 소개

사전에서 키-값 쌍 가져오기

>>> x={'a':1,'b':2}
>>> key,value=x.popitem()
>>> key,value
('a', 1)
>>> del x[key]

Traceback (most recent call last):
  File "<pyshell#16>", line 1, in <module>
    del x[key]
KeyError: &#39;a&#39;
>>> x
{&#39;b&#39;: 2}
>>> x[key]=value
>>> x
{&#39;a&#39;: 1, &#39;b&#39;: 2}
>>> del x[key]

증분 할당

>>> x=2
>>> x+=1
>>> x*=2
>>> x
>>> fnord='foo'
>>> fnord+='bar'
>>> fnord*=2
>>> fnord
'foobarfoobar'

if 문의 조건부 실행

>>> name=raw_input('?')
?Yq Z
>>> if name.endswith('Z'):  \
   print 'Hello,Mr.Z'

Hello,Mr.Z

else 절

>>> name=raw_input('what is your name?')
what is your name?Yq Z
>>> if name.endswith('Z'):
    print 'Hello,Mr.Z'
else:
    print 'Hello,stranger'

    
Hello,Mr.Z

elif 절

>>> num=input('Enter a number: ')
Enter a number: 5
>>> if num>0:
    print 'The number is position'
elif num<0:
    print &#39;The number is negative&#39;
else:
    print &#39;The number is zero&#39;

    
The number is position

조건부 중첩문

>>> name=raw_input('What is your name?')
What is your name?Yq Z
>>> if name.endswith('Yq'):
    if name.startswith('Z'):
        print 'Hello,Yq Z'
    elif name.startswith('K'):
        print 'Hello,Zyq'
    else:
        print 'Hello,Yq'
else:
    print 'Hello,stranger'

    
Hello,stranger

>>> number=input('Enter a number between 1 and 10:')
Enter a number between 1 and 10:6
>>> if number<=10 and number>=1:
    print &#39;Great!&#39;
else:
    print &#39;Wrong!&#39;

    
Great!

>>> age=10
>>> assert 0<age<100
>>> age=-1
>>> assert 0<age<100

Traceback (most recent call last):
  File "<pyshell#21>", line 1, in <module>
    assert 0<age<100
AssertionError

while 루프

>>> x=1
>>> while x<=100:
    print x
    x+=1

>>> while not name:
    name=raw_input('Please enter your name:')
    print 'Hello,%s !' % name

    
Please enter your name:zyq
Hello,zyq !

for 루프

>>> words=['this','is','an','ex','parrot']
>>> for word in words:
    print word

    
this
is
an
ex
parrot
>>> range(0,10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> for i in range(1,8):
    print i
2
4
6

사전 루프(반복)

>>> d={'x':1,'y':2,'z':3}
>>> for key in d:
    print key,'corresponds to',d[key]

    
y corresponds to 2
x corresponds to 1
z corresponds to 3

병렬 반복

>>> names=['Anne','Beth','George','Damon']
  >>> ages=[12,19,18,20]
>>> for i in range(len(names)):
    print names[i],'is',ages[i],'years old'

    
Anne is 12 years old
Beth is 19 years old
George is 18 years old
Damon is 20 years old

rrree

번호가 매겨진 반복

>>> zip(names,ages)
[('Anne', 12), ('Beth', 19), ('George', 18), ('Damon', 20)]
>>> for name,age in zip(names,ages):
    print name,'is',age,'years old'

    
Anne is 12 years old
Beth is 19 years old
George is 18 years old
Damon is 20 years old
>>> zip(range(5),xrange(100))
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)]

뒤집기, 정렬 반복

>>> d
[1, 2, 4, 4]
>>> for x in d:
    if x==4:
        d[d.index(x)]=6

        
>>> d
[1, 2, 6, 6]
>>> S=['skj','kiu','olm','piy']
>>> index=0>>> for s1 in S:
    if 'k' in s1:
        S[index]='HH'
    index+=1

    
>>> S
['HH', 'HH', 'olm', 'piy']>>> for index,s2 in enumerate(S): #enumerate函数提供索引-值对
    if 'H' in s2:
        S[index]='DF'

        
>>> S
['DF', 'DF', 'olm', 'piy']

루프에서 벗어나

>>> sorted([4,3,6,8,3])
[3, 3, 4, 6, 8]
>>> sorted('Hello,world!')
['!', ',', 'H', 'd', 'e', 'l', 'l', 'l', 'o', 'o', 'r', 'w']
>>> list(reversed('Hello,world!'))
['!', 'd', 'l', 'r', 'o', 'w', ',', 'o', 'l', 'l', 'e', 'H']
>>> ''.join(reversed('Hello,world!'))
'!dlrow,olleH'

True/break

>>> for n in range(99,0,-1):
    m=sqrt(n)
    if m==int(m):
        print n
        break

루프

>>> while True:
    word=raw_input('Please enter a word:')
    if not word:break
    print 'The word was '+word

    
Please enter a word:f
The word was f
Please enter a word:

List comprehension-lightweight loop

>>> for n in range(99,81,-1):
    m=sqrt(n)
    if m==int(m):
        print m
        break
else:
    print 'h'

    
h

pass

>>> [x*x for x in range(10)]
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
>>> [x*x for x in range(10) if x%3==0]
[0, 9, 36, 81]
>>> [(x,y) for x in range(3) for y in range (3)]
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
>>> result=[]
>>> for x in range(3):
    for y in range(3):
        result.append((x,y))
>>> result
 [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
>>> girls=['Alice','Bernice','Clarice']
>>> boys=['Chris','Arnold','Bob']
>>> [b+'+'+g for b in boys for g in girls if b[0]==g[0]]
['Chris+Clarice', 'Arnold+Alice', 'Bob+Bernice']

del x와 y의 else 문은 동시에 목록을 가리키지만 x를 삭제하면 y에는 영향을 미치지 않습니다. 목록 자체(값)는 삭제되지 않고 이름만 삭제됩니다.

>>> if name=='Nsds':
    print 'Welcome!'
elif name=='UK':
    #还没完
    pass
elif name=='Bill':
    print 'Access Denied'
else:
    print 'Nobody!'

exec

>>> x=['Hello','world']
>>> y=x
>>> y[1]='Python'
>>> x
['Hello', 'Python']
>>> del x
>>> y
['Hello', 'Python']

참고: 네임스페이스를 범위라고 합니다. 보이지 않는 사전과 마찬가지로 변수를 보관하는 장소라고 생각하세요. x=1과 같은 할당문을 실행하면 키 x와 값 1이 현재 네임스페이스에 배치됩니다. 이 네임스페이스는 일반적으로 전역 네임스페이스입니다.

>>> exec "print 'Hello,world!'"
Hello,world!
>>> from math import sqrt
>>> exec "sqrt=1"
>>> sqrt(4)

Traceback (most recent call last):
  File "<pyshell#36>", line 1, in <module>
    sqrt(4)
TypeError: &#39;int&#39; object is not callable

#增加一个字典，起到命名空间的作用
>>> from math import sqrt
>>> scope={}
>>> exec &#39;sqrt=1&#39; in scope
>>> sqrt(4)
2.0
>>> scope[&#39;sqrt&#39;]

평가는

>>> len(scope)2
>>> scope.keys()
['__builtins__', 'sqrt']

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