如何查找元組中指定列中行的最小值 - Python 2.6？

Question

我在Python 2.6元組，在下面詳細說明：

mylist = [ 
['20120903', 'melon', 'shelf1', '05:31', '08:01'], 
['20120903', 'melon', 'shelf1', '05:31', '14:01'], 
['20120903', 'melon', 'shelf1', '05:31', '23:59'], 
['20120903', 'melon', 'shelf1', '10:18', '14:01'], 
['20120903', 'melon', 'shelf1', '10:18', '23:59'], 
['20120904', 'melon', 'shelf1', '00:00', '14:02'], 
['20120904', 'melon', 'shelf1', '05:32', '14:02'], 
['20120903', 'apple', 'shelf5', '05:34', '14:02'], 
['20120903', 'apple', 'shelf5', '05:34', '23:59'], 
['20120904', 'apple', 'shelf5', '00:00', '14:02'], 
['20120904', 'apple', 'shelf5', '05:33', '14:02']] 

和我想獲得像這些下面的結果（當列0，1，2，3相同，則採取4的最小值個柱+時的列0，1，2，4相同，則取3-RD柱的最大值）：

result = [ 
['20120903', 'melon', 'shelf1', '05:31', '08:01'], 
['20120903', 'melon', 'shelf1', '10:18', '14:01'], 
['20120904', 'melon', 'shelf1', '05:32', '14:02'], 
['20120903', 'apple', 'shelf5', '05:34', '14:02'], 
['20120904', 'apple', 'shelf5', '05:33', '14:02']] 

現在由於阿什維尼喬杜裏，我修改有點他的代碼，現在它如下所示：

from itertools import groupby 

mylist = [ 
['20120903', 'melon', 'shelf1', '05:31', '08:01'], 
['20120903', 'melon', 'shelf1', '05:31', '14:01'], 
['20120903', 'melon', 'shelf1', '05:31', '23:59'], 
['20120903', 'melon', 'shelf1', '10:18', '14:01'], 
['20120903', 'melon', 'shelf1', '10:18', '23:59'], 
['20120904', 'melon', 'shelf1', '00:00', '14:02'], 
['20120904', 'melon', 'shelf1', '05:32', '14:02'], 
['20120903', 'apple', 'shelf5', '05:34', '14:02'], 
['20120903', 'apple', 'shelf5', '05:34', '23:59'], 
['20120904', 'apple', 'shelf5', '00:00', '14:02'], 
['20120904', 'apple', 'shelf5', '05:33', '14:02']] 

step1 = [] 
for k1, g1 in groupby(mylist, key=lambda x1: (x1[0], x1[1], x1[2], x1[3])): 
    step1.append((min(g1, key=lambda x1: map(int, x1[4].split(':'))))) 

step2 = [] 
for k2, g2 in groupby(step1, key=lambda x2: (x2[0], x2[1], x2[2], x2[4])): 
    step2.append((max(g2, key=lambda x2: map(int, x2[3].split(':'))))) 

for result in step2: 
    print result 

Answer 1

我想你需要的東西是這樣的：

>>> from itertools import groupby 
#filter items that contain '00:00' 
>>> mylist = [x for x in mylist if x[-2] != '00:00' ] 

#now group lists based on the the second last item 
for k,g in groupby(mylist, key = lambda x :x [-2]): 
    #find the min among the grouped lists based on the last item 
    minn = min(g, key = lambda x : map(int,x[-1].split(':'))) 
    print minn 
...  
['20120903', 'melon', 'shelf1', '05:31', '08:01'] 
['20120903', 'melon', 'shelf1', '10:18', '14:01'] 
['20120904', 'melon', 'shelf1', '05:32', '14:02'] 
['20120903', 'apple', 'shelf5', '05:34', '14:02'] 
['20120904', 'apple', 'shelf5', '05:33', '14:02'] 

要獲得一個列表的列表，你可以使用發電機功能：

from itertools import groupby 
def solve(lis): 
    mylist = [x for x in lis if x[-2] != '00:00' ] 
    for k,g in groupby(mylist, key = lambda x :x [-2]): 
      #find the min among the grouped lists based on the last item 
      minn = min(g, key = lambda x : map(int,x[-1].split(':'))) 
      yield minn 
...   
>>> list(solve(mylist)) 
[['20120903', 'melon', 'shelf1', '05:31', '08:01'], 
['20120903', 'melon', 'shelf1', '10:18', '14:01'], 
['20120904', 'melon', 'shelf1', '05:32', '14:02'], 
['20120903', 'apple', 'shelf5', '05:34', '14:02'], 
['20120904', 'apple', 'shelf5', '05:33', '14:02']]