嗨我有複雜的數據對象,我想通過s排序。下面的簡化版本:如何使用特定規則對列表進行排序
class Data(object):
def __init__(self, s):
self.s = s
這些數據對象的每一個都將放置在特定的類別中,以方便以後使用。簡體版下面再次
class DataCategory(object):
def __init__(self, id1, id2, linked_data=None):
self.id1 = id1
self.id2 = id2
self.ld = linked_data
我想按照它們的s號碼排序數據但是有更少的規則。如果從第一個數據收集中使用一個數據對象,那麼我想使用第二個數據集中的一個數據對象,如果其數目相同或更低。這裏是我所得到的,我想實現
# order I get
# [['p02g01r05', 5], ['p02g01r01', 4], ['p01g01r05', 4], ['p01g01r01', 3], ['p01g01r02', 2], ['p01g01r03', 2], ['p01g01r06', 2], ['p02g01r02', 2], ['p02g01r03', 2], ['p02g01r04', 2], ['p01g01r04', 1], ['p02g01r06', 1]]
# order I want
# [['p02g01r05', 5], ['p01g01r05', 4], ['p02g01r01', 4], ['p01g01r01', 3], ['p02g01r02', 2], ['p01g01r02', 2], ['p02g01r03', 2], ['p01g01r03', 2], ['p02g01r04', 2], ['p01g01r06', 2], ['p02g01r06', 1]], ['p01g01r04', 1]
這是我創建至今,但我在想,我這個要在錯誤的方向是什麼。我認爲,要替換的索引列表是正確的。
# Some data objects
p01g01r01 = Data(3)
p01g01r02 = Data(2)
p01g01r03 = Data(2)
p01g01r04 = Data(1)
p01g01r05 = Data(4)
p01g01r06 = Data(2)
p02g01r01 = Data(4)
p02g01r02 = Data(2)
p02g01r03 = Data(2)
p02g01r04 = Data(2)
p02g01r05 = Data(5)
p02g01r06 = Data(1)
p01g01 = DataCategory("01", "01", [])
p02g01 = DataCategory("02", "01", [])
# link data to data category
def ldtdc(dc):
lst = []
data = "p" + dc.id1 + "g" + dc.id2 + "r"
for i in range(1, 7):
if i < 10:
lst.append(data + "0" + str(i))
else:
lst.append(data + str(i))
return lst
p01g01.ld = ldtdc(p01g01)
p02g01.ld = ldtdc(p02g01)
# /@= This starts to get way too complicated fast ############################
def lstu(ag, dg):
lst = []
# data list of first collection
dlofc = []
# data list of second collection
dlosc = []
# for every data unit that exists in data collection
for unit in ag.ld:
# lst.append([unit, globals()[unit].s+10])
lst.append([unit, globals()[unit].s])
dlofc.append([unit, globals()[unit].s])
for unit in dg.ld:
lst.append([unit, globals()[unit].s])
dlosc.append([unit, globals()[unit].s])
# lambda function is used here to sort list by data value ([1] is index of the item)
lst = sorted(lst, key=lambda x: x[1], reverse=True)
# current index
ci = 0
previous_data = ["last data unit will be stored here", 0]
# sorted list
slst = []
for unit in lst:
try:
next_data = lst[ci+1]
except IndexError:
next_data = ["endoflist", 0]
if previous_data[0] == "last data unit will be stored here":
pass
elif previous_data[0][:6] == unit[0][:6]:
if unit[0][:6] not in dlofc[0][0]:
slst.append([unit[0], unit[1], ci])
elif unit[0][:6] not in dlosc[0][0]:
slst.append([unit[0], unit[1], ci])
else:
print "Error"
previous_data = unit
ci += 1
print "slist below"
print slst
return lst
# \@= END #####################################################################
print p01g01.ld
print p02g01.ld
data_list = lstu(p01g01, p02g01)
print data_list
什麼是排序這種數據的快速和正確的方法?
你考慮過'sorted'函數或'list.sort'方法嗎? – skyking
在上面的例子中,你可以看到我已經使用了排序,但它不足以滿足新列表的所有要求。 – Hsin
你知道/意識到你可以控制'sorted'和'list.sort'在排序時比較元素的方式?一旦你可以控制,我不明白你爲什麼不應該能夠使用'sorted'或'list.sort'。 – skyking