Question

示例:

help([[1, 2, 2], [1, 1, 2]])

返回 2。

def help(num):
    lst = ()
    for n in num:  
        return(len(lst))

所以我需要桌上所有数字的总和

另一个例子:

help([[32, 12, 52, 63], [32, 64, 67, 52], [64, 64, 17, 34], [34, 17, 76, 32]])

返回: 9

Answer 1

def allElementsInTable(table):
    for row in table:
        for elem in row:
            yield elem

演示 :

>>> set(allElementsInTable([[1,2,2],[1,1,2]]))
{1,2}

>>> len(set(allElementsInTable([[1,2,2],[1,1,2]])))
2

如果您做了很多表格操作操作, 您可能想要查看 < code> numpy 数组。

Answer 2

试试这个 (< a href=>"http://docs.python.org/library/itertools.html#itertools. lex" rel="nofollow">itertools. 链用于“ contailate” the exterables):

table = [ [32, 12, 52, 63], 
          [32, 64, 67, 52], 
          [64, 64, 17, 34], 
          [34, 17, 76, 32] ]

from itertools import chain
n_unique_values = len(set( chain(*table) ))

print n_unique_values # prints 9

Answer 3

>>> x = [[1, 2, 2], [1, 1, 2]]
>>> unique_items = lambda n: len(set(n))
>>> map(unique_items, x)
[2, 2]

Answer 4

@garnertb当然是正确的,

>>> x = [[1, 2, 2], [1, 1, 2]]
>>> map(len, map(set, x))
[2, 2]

问题在于如果x变长,

>>> import itertools
>>> itertools.imap(len, itertools.imap(set, x))
<itertools.imap object at 0xb7437b8c>

Answer 5

compound_list = []
for sublist in lists:
    compound_list += sublist
len(set(compound_list))

友情链接