判断列表是否为空
if not a:
print "List is empty"
这里是因为空列表会返回一个False
获取列表的索引和值
ints = [8, 23, 45, 12, 78]
for idx, val in enumerate(ints):
print idx, val
0 8
1 23
2 45
3 12
4 78
合并列表中的子列表
1 method
l=[[1,2,3],[4,5,6], [7], [8,9]]
print([item for sublist in l for item in sublist])
[1, 2, 3, 4, 5, 6, 7, 8, 9]
2 method
l=[[1,2,3],[4,5,6], [7], [8,9]]
print(sum(l, []))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
3 method
from functools import reduce
l=[[1,2,3],[4,5,6], [7], [8,9]]
print(reduce(lambda x,y: x+y,l))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
第一种方法是速度最快的方法
列表中字典的值排序
list_to_be_sorted = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
from operator import itemgetter
newlist = sorted(list_to_be_sorted, key=itemgetter('name'))
print(newlist)
把列表分割成同样大小的块
tuple(l[i:i+n] for i in xrange(0, len(l), n))
def chunks(l, n):
""" Yield successive n-sized chunks from l.
"""
for i in xrange(0, len(l), n):
yield l[i:i+n]
合并两个列表
mergedlist = listone + listtwo
列表中随机取一个元素
foo = ['a', 'b', 'c', 'd', 'e']
print(random.choice(foo))
按照步长遍历列表
a[start:end:step] # 按照step步长直到end-1结束,并不是从start一个个遍历到end
Python中的appen和extend
x = [1, 2, 3]
x.append([4, 5])
print (x)
输出:[1, 2, 3, [4, 5]]
x = [1, 2, 3]
x.extend([4, 5])
print (x)
输出:[1, 2, 3, 4, 5]
