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從條件列表中刪除列表

[英]Removing List from List of Lists with condition

 原文  2016-09-08 02:39:17  5   3    python/ list/ conditional-statements

問題描述

我有這個項目列表，我想從最后一個項目到項目n一個接一個地刪除它，直到它達到預算的總值= 325.000 

from collections import namedtuple

Item = namedtuple('Item', 'region sector name budget target performance'.split())

sorted_KP = [Item(region='H', sector='2', name='H3', budget=7000.0, target=1.0, performance=4.0),
Item(region='H', sector='2', name='H10', budget=35000.0, target=15.0, performance=1.0),
Item(region='I', sector='2', name='I6', budget=50000.0, target=5.0, performance=0.40598931548848194),
Item(region='E', sector='4', name='E5', budget=75000.0, target=30.0, performance=0.0663966081766),
Item(region='C', sector='1', name='C1', budget=75000.0, target=50.0, performance=0.0308067750379),
Item(region='C', sector='1', name='C2', budget=75000.0, target=50.0, performance=0.0308067750379),
Item(region='C', sector='5', name='C4', budget=75000.0, target=50.0, performance=0.0308067750379),
Item(region='I', sector='2', name='I5', budget=100000.0, target=5.0, performance=0.40598931548848194),
Item(region='E', sector='4', name='E1', budget=100000.0, target=30.0, performance=0.0663966081766),
Item(region='D', sector='5', name='D21', budget=60000.0, target=4.0, performance=0.2479775110248),
Item(region='D', sector='5', name='D30', budget=10000.0, target=1.0, performance=0.1653183406832),
Item(region='D', sector='1', name='D23', budget=30000.0, target=20.0, performance=0.023659703723372342),
Item(region='C', sector='5', name='C3', budget=150000.0, target=75.0, performance=0.0308067750379),
Item(region='D', sector='5', name='D20', budget=30000.0, target=5.0, performance=0.0826591703416),
Item(region='H', sector='2', name='H6', budget=310576.0, target=1.0, performance=4.0),
Item(region='H', sector='3', name='H5', budget=9500.0, target=1.0, performance=0.1172008400616),
Item(region='E', sector='6', name='E3', budget=100000.0, target=30.0, performance=0.03747318294316411),
Item(region='G', sector='3', name='G17', budget=75000.0, target=20.0, performance=0.04132095963602382),
Item(region='C', sector='4', name='C5', budget=75000.0, target=25.0, performance=0.0308067750379),
Item(region='C', sector='2', name='C6', budget=30000.0, target=5.0, performance=0.0616135500758),
Item(region='C', sector='2', name='C7', budget=30000.0, target=5.0, performance=0.0616135500758),
Item(region='D', sector='6', name='D22', budget=65190.0, target=30.0, performance=0.020332158889648923),
Item(region='D', sector='5', name='D3', budget=100000.0, target=20.0, performance=0.0413295851708),
Item(region='D', sector='5', name='D4', budget=100000.0, target=20.0, performance=0.0413295851708),
Item(region='A', sector='1', name='A12', budget=25000.0, target=25.0, performance=0.00749432996938),
Item(region='A', sector='1', name='A13', budget=25000.0, target=25.0, performance=0.00749432996938),
Item(region='A', sector='3', name='A25', budget=4500.0, target=1.0, performance=0.02997731987752),
Item(region='A', sector='5', name='A26', budget=4500.0, target=1.0, performance=0.02997731987752),
Item(region='A', sector='1', name='A27', budget=4500.0, target=1.0, performance=0.02997731987752),
Item(region='A', sector='1', name='A29', budget=4500.0, target=1.0, performance=0.02997731987752),
Item(region='A', sector='3', name='A30', budget=4500.0, target=1.0, performance=0.02997731987752)]

但除了總價值之外，我還有另外兩個項目的條件是否應該刪除。

首先，在刪除項目之后，仍然存在代表相同區域的列表列表中的至少一個項目

其次，在移除項目之后，仍然存在至少一個表示相同扇區的項目

例如，我可以刪除最后一項，因為它代表區域“A”，剩下5項也代表區域“A”。 它也代表扇區“3”，留下3個代表扇區“3”的項目。

重復此刪除和檢查，直到我達到總移除預算至少325.000

我做了這個代碼，但我無法得到我需要的東西。 請幫我糾正一下。 

from collections import Counter
unpack = []
for item in sorted_KP:
    item_budget = item[3]
    sum_unpack = sum(item[3] for item in unpack)
    budget = 325000

    remaining = []
    for item in sorted_KP:
         if item not in unpack:
             remaining.append(item)

    region_el = [item[0] for item in remaining]
    counter_R_el = Counter(region_el)

    sector_el = [item[1] for item in remaining]
    counter_S_el = Counter(sector_el)

    if counter_R_el >= 1 or counter_S_el >= 1:
        if sum_unpack <= budget:
            unpack.append(item)

for item in unpack:
    print "\t", item

這是我的代碼，第25項仍然刪除時，它不應該： 

unpack =Item(region='A', sector='3', name='A30', budget=4500.0, target=1.0, performance=0.02997731987752)
    Item(region='A', sector='1', name='A29', budget=4500.0, target=1.0, performance=0.02997731987752)
    Item(region='A', sector='1', name='A27', budget=4500.0, target=1.0, performance=0.02997731987752)
    Item(region='A', sector='5', name='A26', budget=4500.0, target=1.0, performance=0.02997731987752)
    Item(region='A', sector='3', name='A25', budget=4500.0, target=1.0, performance=0.02997731987752)
    Item(region='A', sector='1', name='A13', budget=25000.0, target=25.0, performance=0.00749432996938)
    Item(region='A', sector='1', name='A12', budget=25000.0, target=25.0, performance=0.00749432996938)
    Item(region='D', sector='5', name='D4', budget=100000.0, target=20.0, performance=0.0413295851708)
    Item(region='D', sector='5', name='D3', budget=100000.0, target=20.0, performance=0.0413295851708)
    Item(region='D', sector='6', name='D22', budget=65190.0, target=30.0, performance=0.020332158889648923)

項目25（項目名稱：“A12”）無法刪除，即使我們仍有預算要刪除，因為如果刪除，則不再有項目代表區域“A”，依此類推。

雖然解決方案應該是： 

unpack = [Item(region='A', sector='3', name='A30', budget=4500.0, target=1.0, performance=0.02997731987752),
Item(region='A', sector='1', name='A29', budget=4500.0, target=1.0, performance=0.02997731987752),
Item(region='A', sector='1', name='A27', budget=4500.0, target=1.0, performance=0.02997731987752),
Item(region='A', sector='5', name='A26', budget=4500.0, target=1.0, performance=0.02997731987752),
Item(region='A', sector='3', name='A25', budget=4500.0, target=1.0, performance=0.02997731987752),
Item(region='A', sector='1', name='A13', budget=25000.0, target=25.0, performance=0.00749432996938),
Item(region='D', sector='5', name='D4', budget=100000.0, target=20.0, performance=0.0413295851708),
Item(region='D', sector='5', name='D3', budget=100000.0, target=20.0, performance=0.0413295851708),
Item(region='D', sector='6', name='D22', budget=65190.0, target=30.0, performance=0.020332158889648923),
Item(region='C', sector='2', name='C7', budget=30000.0, target=5.0, performance=0.0616135500758)]

預先感謝您的幫助

3 個解決方案

解決方案1
 1  2016-09-08 04:09:53

在我實際嘗試運行它時更新答案我發現了一些其他問題：

  • for item in sorted_KP使用與外部循環相同的item計數器並覆蓋它 - 始終嘗試刪除A30 （最后）項目
  • 當切換到內循環中的item2 ，我還必須反轉外循環順序（即從最后一行開始刪除）。
  • 區域/扇區計數器比較不正確，導致TypeError: unorderable types: Counter() >= int() - 需要根據需要選擇匹配項目區域或扇區的特定計數
  • 結合我之前的回答：你需要and你的2個額外條件，不是or他們
  • 並入@wwii的評論 - 實際上，計數器比較需要> 1 ，而不是>= 1

實際測試的代碼： 

>>> for item in sorted_KP[::-1]:
...     item_budget = item[3]
...     sum_unpack = sum(item[3] for item in unpack)
...     budget = 325000
...     remaining = []
...     for item2 in sorted_KP:
...          if item2 not in unpack:
...              remaining.append(item2)
...     region_el = [item[0] for item in remaining]
...     counter_R_el = Counter(region_el)
...     sector_el = [item[1] for item in remaining]
...     counter_S_el = Counter(sector_el)
...     if counter_R_el[item.region] > 1 and counter_S_el[item.sector] > 1:
...         if sum_unpack <= budget:
...             unpack.append(item)
... 
>>> 
>>> for item in unpack:
...    logging.error(item)
... 
ERROR:root:Item(region='A', sector='3', name='A30', budget=4500.0, target=1.0, performance=0.02997731987752)
ERROR:root:Item(region='A', sector='1', name='A29', budget=4500.0, target=1.0, performance=0.02997731987752)
ERROR:root:Item(region='A', sector='1', name='A27', budget=4500.0, target=1.0, performance=0.02997731987752)
ERROR:root:Item(region='A', sector='5', name='A26', budget=4500.0, target=1.0, performance=0.02997731987752)
ERROR:root:Item(region='A', sector='3', name='A25', budget=4500.0, target=1.0, performance=0.02997731987752)
ERROR:root:Item(region='A', sector='1', name='A13', budget=25000.0, target=25.0, performance=0.00749432996938)
ERROR:root:Item(region='D', sector='5', name='D4', budget=100000.0, target=20.0, performance=0.0413295851708)
ERROR:root:Item(region='D', sector='5', name='D3', budget=100000.0, target=20.0, performance=0.0413295851708)
ERROR:root:Item(region='D', sector='6', name='D22', budget=65190.0, target=30.0, performance=0.020332158889648923)
ERROR:root:Item(region='C', sector='2', name='C7', budget=30000.0, target=5.0, performance=0.0616135500758)
>>>

解決方案2
 1 已采納  2016-09-08 08:47:01

假設您在使用print without ()使用Python 2。

請參閱修改后的代碼中的注釋： 

unpack = []
for item in sorted_KP[::-1]: # loop the items in reverse order
    #item_budget = item[3] # variable not used
    sum_unpack = sum(item[3] for item in unpack)
    budget = 325000

    remaining = []
    for x in sorted_KP: # don't use 'item' here as in Python 2, it will modify the variable 'item' in the outer loop
        if x not in unpack:
            remaining.append(x)

    region_el = [x[0] for x in remaining] # don't use 'item' here as well
    counter_R_el = Counter(region_el)

    sector_el = [x[1] for x in remaining] # don't use 'item' here as well
    counter_S_el = Counter(sector_el)

    #if counter_R_el >= 1 or counter_S_el >= 1: # note that result is always True in Python 2
    if counter_R_el[item.region] > 1 and counter_S_el[item.sector] > 1: # check '> 1' instead of '>= 1', and use 'and' instead of 'or'
        if sum_unpack <= budget:
            unpack.append(item)

for item in unpack:
    print "\t", item

我建議的解決方案 

budget = 325000
sum = 0 # accumulated budget of removed items
removed_KP = [] # holds the removed items
for item in sorted_KP[::-1]:
    # stop checking if over-budget
    if sum >= budget: break
    # check whether there are items with same region and sector
    rcnt = scnt = 0
    for tmp in sorted_KP:
        if tmp.region == item.region: rcnt += 1
        if tmp.sector == item.sector: scnt += 1
    if scnt > 1 and rcnt > 1:
        # this item can be removed based on the constraints
        sorted_KP.remove(item)
        removed_KP.append(item)
        sum += item.budget

# print the removed items
for item in removed_KP:
    print(item)

解決方案3
 0  2016-09-08 05:41:33

我無法輕松整理您的代碼。 if counter_R_el >= 1 or counter_S_el >= 1:只是無法工作，則無法將Counter（）與int（）進行比較。 它看起來像你不斷重制/重置你的循環套件中的東西 ，它變得令人困惑。

你似乎走在了正確的軌道上。

  • 要跟蹤區域和行業集合的數量.Counter（）是一個不錯的選擇。
  • 查看列表中的每個項目，並檢查它是否符合您的約束
  • 如果刪除，請更新計數器和總費用

這是我想出的： 

import collections, operator
Item = collections.namedtuple('Item', ['region', 'sector', 'name',
                                       'budget', 'target', 'performance'])

a = [Item(region='H', sector='2', name='H3', budget=7000.0, target=1.0, performance=4.0),
     Item(region='H', sector='2', name='H10', budget=35000.0, target=15.0, performance=1.0),
     Item(region='I', sector='2', name='I6', budget=50000.0, target=5.0, performance=0.40598931548848194),
     .....]

budget = 325000

# sort highest to lowest cost
a.sort(key = operator.attrgetter('budget'), reverse = True)
project_cost = sum(item.budget for item in a)
# constraint counters
region_count = collections.Counter(item.region for item in a)
sector_count = collections.Counter(item.sector for item in a)

# iterate over a copy of the list
b = a.copy()
for item in b:
    if project_cost <= budget:
        break
    # check item against constraints
    if region_count[item.region] > 1 and sector_count[item.sector] > 1:
        # remove the item from the original list 
        a.remove(item)
        # uodate the counters and cost
        region_count[item.region] -= 1
        sector_count[item.sector] -= 1
        project_cost -= item.budget

a包含符合約束條件且總成本低於預算的項目。

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