我尝试总结嵌套元素列表
例如，numbers=[1,3,5,6,[7,8]]应该产生sum=30
我写了以下代码：
def nested_sum(L):
sum=0
for i in range(len(L)):
if (len(L[i])>1):
sum=sum+nested_sum(L[i])
else:
sum=sum+L[i]
return sum
上面的代码给出了以下错误：
'int' 类型的对象没有 len()
我也试过了len([L[i]])，还是不行。
任何人都可以帮忙吗？它是 Python 3.3
您需要使用isinstance来检查元素是否为列表。此外，您可能希望遍历实际列表，以使事情变得更简单。
def nested_sum(L):
total = 0 # don't use `sum` as a variable name
for i in L:
if isinstance(i, list): # checks if `i` is a list
total += nested_sum(i)
else:
total += i
return total
通常认为鸭类型更 Pythonic ，而不是显式类型检查。像这样的东西将采用任何可迭代的，而不仅仅是列表：
def nested_sum(a) :
total = 0
for item in a :
try:
total += item
except TypeError:
total += nested_sum(item)
return total
具有列表理解的另一种解决方案：
>>> sum( sum(x) if isinstance(x, list) else x for x in L )
30
编辑：对于超过两个级别的列表（谢谢@Volatility）：
def nested_sum(L):
return sum( nested_sum(x) if isinstance(x, list) else x for x in L )
我会总结扁平列表：
def flatten(L):
'''Flattens nested lists or tuples with non-string items'''
for item in L:
try:
for i in flatten(item):
yield i
except TypeError:
yield item
>>> sum(flatten([1,3,5,6,[7,8]]))
30
使用 lambda 处理嵌套列表的快速递归：
rec = lambda x: sum(map(rec, x)) if isinstance(x, list) else x
rec，应用于列表，将返回总和（递归），在一个值上，返回该值。
result = rec(a)
使用过滤器和映射以及递归的示例：
def islist(x):
return isinstance(x, list)
def notlist(x):
return not isinstance(x, list)
def nested_sum(seq):
return sum(filter(notlist, seq)) + map(nested_sum, filter(islist, seq))
这是一个使用reduce和recursion的例子
from functools import reduce
def nested_sum(seq):
return reduce(lambda a,b: a+(nested_sum(b) if isinstance(b, list) else b), seq)
使用普通旧递归的示例：
def nested_sum(seq):
if isinstance(seq[0], list):
head = nested_sum(seq[0])
else:
head = seq[0]
return head + nested_sum(seq[1:])
使用模拟递归的示例：
def nested_sum(seq):
stack = []
stack.append(seq)
result = 0
while stack:
item = stack.pop()
if isinstance(item, list):
for e in item:
stack.append(e)
else:
result += item
return result
处理自引用列表的调整留给读者作为练习。
此代码也有效。
def add_all(t):
total = 0
for i in t:
if type(i) == list: # check whether i is list or not
total = total + add_all(i)
else:
total += i
return total
def sum_nest_lst(lst):
t=0
for l in lst:
if(type(l)==int):
t=t+l
if(type(l)==list):
t=t+sum(l)
print(t)
def nnl(nl): # non nested list function
nn = []
for x in nl:
if type(x) == type(5):
nn.append(x)
if type(x) == type([]):
n = nnl(x)
for y in n:
nn.append(y)
return sum(nn)
print(nnl([[9, 4, 5], [3, 8,[5]], 6])) # output:[9,4,5,3,8,5,6]
a = sum(nnl([[9, 4, 5], [3, 8,[5]], 6]))
print (a) # output: 40
一个简单的解决方案是使用嵌套循环。
def nested_sum(t):
sum=0
for i in t:
if isinstance(i, list):
for j in i:
sum +=j
else:
sum += i
return sum
L = [1, 2, 3, [4, 5, 6], 5, [7, 8, 9]]
total = 0 # assign any var
for a in L: # assign index and start to iterate using if else
if (isinstance(a, list)): # since its a list you are basically repeating the prev step
for b in a:
total += b
else:
total += a
print(total)
def list_sum(L):
return sum(list_sum(x) if isinstance(x, list) else x for x in L)
def nested_sum(lists):
total = 0
for lst in lists:
s = sum(lst)
total += s
return total
#nested sum
l = [[1, 2], [3,5], [6,2], [4, 5, 6,9]]
def nested_sum(lst):
sum = 0
for i in lst:
for j in i:
sum = sum + j
print(sum)
nested_sum(l)