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#字符串遍历 str='aaabbbbccc' for i in str: print(i) #列表遍历 l=['1','a','b'] #方法1 遍历值 for i in l: print(i) #方法2 索引位置和对应值都遍历出来,此方法还可以应用在元祖与字符串上 for k,v in enumerate(l): print(k,v) #元祖遍历 t=(1,'aa','哈') for i in t: print(i) #字典遍历 d={1:'a',2:'b',3:'c'} #方法1 for i in d: print(i,d[i]) #方法2 for k,v in d.items(): print(k,v) #集合遍历 se={'a','b',1,2} for i in se: print(i)
t=(1,'aa','哈')
for i in reversed(t):
print(i)
questions = ['name', 'quest', 'favorite color']
answers = ['lancelot', 'the holy grail', 'blue']
for q, a in zip(questions, answers):
print('What is your {0}? It is {1}.'.format(q, a))
延伸学习zip()打包
a = [1,2,3]
b = [4,5,6]
c = [4,5,6,7,8]
zipped = zip(a,b) # 打包为元组的列表
#结果:[(1, 4), (2, 5), (3, 6)]
zip(a,c)# # 元素个数与最短的列表一致
#结果:[(1, 4), (2, 5), (3, 6)]
#如何遍历列表中的列表
def l(a):
for i in a:
if isinstance(i,(str,int)):
print(i)
else:
l(i)#直接利用函数来递归循环
if __name__=='__main__':
a = [[1, [1, 2]], 'cc', 3, ['a', 'b'],('哈哈哈','hhh')]
l(a)
结果
1
1
2
cc
3
a
b
哈哈哈
hhh
#递归遍历字典
def list_dictionary(d, n_tab=-1): if isinstance(d, list): for i in d: list_dictionary(i, n_tab) elif isinstance(d, dict): n_tab+=1 for key, value in d.items(): print("{}key:{}".format("\t"*n_tab, key)) list_dictionary(value, n_tab) else: print("{}{}".format("\t"*n_tab, d)) if __name__=='__main__': a = [[1, [1, 2]], 'cc', 3, ['a', 'b'],('哈哈哈','hhh'),{99:'goos'}] list_dictionary(a) b={'dd':'3333','age':{1:'a',2:'b'},'old':{3:"c",4:'d'},'cl':{5:'gg',8:"哈哈"}} list_dictionary(b)
快速将列表的内容转为字符串
a=["aa","ddd"]
print(",".join(a))
方法一:
a='a'
b=['a','b']
c=list(map(lambda x: x+a,b))# 使用 lambda 匿名函数
print(c)
方法二
c=['a'+x for x in b]#列表前面加某一个字符
print(c)
结果:[‘aa’, ‘ab’]
a=[1,2]
c=[str(i) for i in a ]
print(c)
结果
[‘1’, ‘2’]
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