排序和在Python

Question

分組嵌套列表

我有以下數據結構（列表的列表）

[ 
['4', '21', '1', '14', '2008-10-24 15:42:58'], 
['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
['5', '21', '3', '19', '2008-10-24 15:45:45'], 
['6', '21', '1', '1somename', '2008-10-24 15:45:49'], 
['7', '22', '3', '2somename', '2008-10-24 15:45:51'] 
] 

我希望能夠

1. 使用重新排序列表功能以便我可以按列表中的每個項目進行分組。例如，我希望能夠按第二列進行分組（所有21都在一起）

2. 使用函數僅顯示每個內部列表中的某些值。例如，我想，以減少該列表只包含「2somename」第四屆字段值

所以列表看起來像這樣

[ 
    ['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
    ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] 
] 

Answer 1

工作對於第一個問題，首先映入你的第4個字段值應該做的是排序列表中第二場：

x = [ 
['4', '21', '1', '14', '2008-10-24 15:42:58'], 
['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
['5', '21', '3', '19', '2008-10-24 15:45:45'], 
['6', '21', '1', '1somename', '2008-10-24 15:45:49'], 
['7', '22', '3', '2somename', '2008-10-24 15:45:51'] 
] 

from operator import itemgetter 

x.sort(key=itemgetter(1)) 

然後你可以使用itertools' GROUPBY功能：

from itertools import groupby 
y = groupby(x, itemgetter(1)) 

現在y是一個包含（元素，項目迭代器）元組的迭代器。這是較爲混亂來解釋這些元組比它表明代碼：

for elt, items in groupby(x, itemgetter(1)): 
    print(elt, items) 
    for i in items: 
     print(i) 

它打印：

21 <itertools._grouper object at 0x511a0> 
['4', '21', '1', '14', '2008-10-24 15:42:58'] 
['5', '21', '3', '19', '2008-10-24 15:45:45'] 
['6', '21', '1', '1somename', '2008-10-24 15:45:49'] 
22 <itertools._grouper object at 0x51170> 
['3', '22', '4', '2somename', '2008-10-24 15:22:03'] 
['7', '22', '3', '2somename', '2008-10-24 15:45:51'] 

對於第二部分，你應該使用列表內涵提到已經在這裏：

from pprint import pprint as pp 
pp([y for y in x if y[3] == '2somename']) 

哪打印：

[['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
['7', '22', '3', '2somename', '2008-10-24 15:45:51']] 

Answer 2

如果我正確的理解你的問題下面的代碼應該做的工作：

l = [ 
['4', '21', '1', '14', '2008-10-24 15:42:58'], 
['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
['5', '21', '3', '19', '2008-10-24 15:45:45'], 
['6', '21', '1', '1somename', '2008-10-24 15:45:49'], 
['7', '22', '3', '2somename', '2008-10-24 15:45:51'] 
] 

def compareField(field): 
    def c(l1,l2): 
     return cmp(l1[field], l2[field]) 
    return c 

# Use compareField(1) as the ordering criterion, i.e. sort only with 
# respect to the 2nd field 
l.sort(compareField(1)) 
for row in l: print row 

print 
# Select only those sublists for which 4th field=='2somename' 
l2somename = [row for row in l if row[3]=='2somename'] 
for row in l2somename: print row 

輸出：

['4', '21', '1', '14', '2008-10-24 15:42:58'] 
['5', '21', '3', '19', '2008-10-24 15:45:45'] 
['6', '21', '1', '1somename', '2008-10-24 15:45:49'] 
['3', '22', '4', '2somename', '2008-10-24 15:22:03'] 
['7', '22', '3', '2somename', '2008-10-24 15:45:51'] 

['3', '22', '4', '2somename', '2008-10-24 15:22:03'] 
['7', '22', '3', '2somename', '2008-10-24 15:45:51'] 

Answer 3

如果你把它分配給VAR 「一」 ...

＃1：

a.sort(lambda x,y: cmp(x[1], y[1])) 

＃2：

filter(lambda x: x[3]=="2somename", a) 

Answer 4

使用函數重新排序列表，以便我可以按列表中的每個項目進行分組。例如，我希望能夠按第二列進行分組（所有21都在一起）

列表有一個內置的排序方法，您可以提供一個提取排序鍵的函數。

>>> import pprint 
>>> l.sort(key = lambda ll: ll[1]) 
>>> pprint.pprint(l) 
[['4', '21', '1', '14', '2008-10-24 15:42:58'], 
['5', '21', '3', '19', '2008-10-24 15:45:45'], 
['6', '21', '1', '1somename', '2008-10-24 15:45:49'], 
['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
['7', '22', '3', '2somename', '2008-10-24 15:45:51']] 

使用僅顯示從每個內部列表中的某些值的函數。例如，我想，以減少該列表只包含「2somename」

這看起來像list comprehensions

>>> [ll[3] for ll in l] 
['14', '2somename', '19', '1somename', '2somename'] 

Answer 5

如果您要進行大量排序和過濾，您可能會喜歡一些幫助功能。

m = [ 
['4', '21', '1', '14', '2008-10-24 15:42:58'], 
['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
['5', '21', '3', '19', '2008-10-24 15:45:45'], 
['6', '21', '1', '1somename', '2008-10-24 15:45:49'], 
['7', '22', '3', '2somename', '2008-10-24 15:45:51'] 
] 

# Sort and filter helpers. 
sort_on = lambda pos:  lambda x: x[pos] 
filter_on = lambda pos,val: lambda l: l[pos] == val 

# Sort by second column 
m = sorted(m, key=sort_on(1)) 

# Filter on 4th column, where value = '2somename' 
m = filter(filter_on(3,'2somename'),m) 

Answer 6

看起來很像你試圖使用列表作爲數據庫。

當今Python在覈心發行版中包含sqlite綁定。如果您不需要持久性，那麼創建內存中的sqlite數據庫非常簡單（請參閱How do I create a sqllite3 in-memory database?）。

然後，您可以使用SQL語句來執行所有這些排序和過濾，而無需重新發明輪子。

Answer 7

對於部分（2），其中x爲您的數組，我想你想，

[y for y in x if y[3] == '2somename'] 

將返回具有第四值是「2somename」只是你的數據列表的列表...儘管看起來卡米爾正在爲SQL提供最好的建議......

Answer 8

你只是在你的結構上創建索引，對不對？

>>> from collections import defaultdict 
>>> def indexOn(things, pos): 
...  inx= defaultdict(list) 
...  for t in things: 
...    inx[t[pos]].append(t) 
...  return inx 
... 
>>> a=[ 
... ['4', '21', '1', '14', '2008-10-24 15:42:58'], 
... ['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
... ['5', '21', '3', '19', '2008-10-24 15:45:45'], 
... ['6', '21', '1', '1somename', '2008-10-24 15:45:49'], 
... ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] 
... ] 

這是你的第一個請求，按位置分組1.

>>> import pprint 
>>> pprint.pprint(dict(indexOn(a,1))) 
{'21': [['4', '21', '1', '14', '2008-10-24 15:42:58'], 
     ['5', '21', '3', '19', '2008-10-24 15:45:45'], 
     ['6', '21', '1', '1somename', '2008-10-24 15:45:49']], 
'22': [['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
     ['7', '22', '3', '2somename', '2008-10-24 15:45:51']]} 

這裏是你的第二個請求，通過位置3.

>>> dict(indexOn(a,3)) 
{'19': [['5', '21', '3', '19', '2008-10-24 15:45:45']], '14': [['4', '21', '1', '14', '2008-10-24 15:42:58']], '2somename': [['3', '22', '4', '2somename', '2008-10-24 15:22:03'], ['7', '22', '3', '2somename', '2008-10-24 15:45:51']], '1somename': [['6', '21', '1', '1somename', '2008-10-24 15:45:49']]} 
>>> pprint.pprint(_) 
{'14': [['4', '21', '1', '14', '2008-10-24 15:42:58']], 
'19': [['5', '21', '3', '19', '2008-10-24 15:45:45']], 
'1somename': [['6', '21', '1', '1somename', '2008-10-24 15:45:49']], 
'2somename': [['3', '22', '4', '2somename', '2008-10-24 15:22:03'], 
       ['7', '22', '3', '2somename', '2008-10-24 15:45:51']]}