在python列表中保留日期順序時插入缺少的日期

Question

我有一個包含[yyyy，value]項目的列表列表，每個子列表按遞增年份排序。 這是一個示例：

A = [
    [[2008, 5], [2009, 5], [2010, 2], [2011, 5], [2013, 17]], 
    [[2008, 6], [2009, 3], [2011, 1], [2013, 6]], [[2013, 9]], 
    [[2008, 4], [2011, 1], [2013, 4]], 
    [[2010, 3], [2011, 3], [2013, 1]], 
    [[2008, 2], [2011, 4], [2013, 1]], 
    [[2009, 1], [2010, 1], [2011, 3], [2013, 3]], 
    [[2010, 1], [2011, 1], [2013, 5]], 
    [[2011, 1], [2013, 4]], 
    [[2009, 1], [2013, 4]], 
    [[2008, 1], [2013, 3]], 
    [[2009, 1], [2013, 2]], 
    [[2013, 2]], 
    [[2011, 1], [2013, 1]],
    [[2013, 1]], 
    [[2013, 1]], 
    [[2011, 1]], 
    [[2011, 1]]
    ]

我需要的是在min（year）和max（year）之間插入所有缺失的年份，並確保保留訂單。 因此，例如，取A的第一個子列表：

[2008, 5], [2009, 5], [2010, 2], [2011, 5], [2013, 17]

應該是這樣的：

[min_year, 0]...[2008, 5], [2009, 5], [2010, 2], [2011, 5], [2012, 0],[2013, 17],..[max_year, 0]

此外，如果任何子列表僅包含單個項目，則應對其應用相同的過程，以便原始值保留其假定的順序，並且正確插入min到max（year，value）項目的其余部分。

有任何想法嗎？

謝謝。

Answer 1

怎么樣：

import numpy as np

def np_fill(data,min_year,max_year):

    #Setup empty array
    year_range=np.arange(min_year,max_year+1)
    unit=np.dstack((year_range,np.zeros(max_year-min_year+1)))
    overall=np.tile(unit,(len(data),1,1)).astype(np.int)

    #Change the list to a list of ndarrays
    data=map(np.array,data)

    for num,line in enumerate(data):

        #Find correct indices and update overall array
        index=np.searchsorted(year_range,line[:,0])
        overall[num,index,1]=line[:,1]
    return overall

運行代碼：

print np_fill(A,2008,2013)[:2]

[[[2008    5]
  [2009    5]
  [2010    2]
  [2011    5]
  [2012    0]
  [2013   17]]

 [[2008    6]
  [2009    3]
  [2010    0]
  [2011    1]
  [2012    0]
  [2013    6]]]


print np_fill(A,2008,2013).shape
(18, 6, 2)

您在A的第二行中有2013年的副本，不確定這是否有目的。

有幾個時間因為我很好奇，源代碼可以在這里找到。 如果您發現錯誤，請告訴我。

開始年/年 - （2008,2013）：

np_fill took 0.0454630851746 seconds.
tehsockz_fill took 0.00737619400024 seconds.
zeke_fill_fill took 0.0146050453186 seconds.

有點期待這一點 - 轉換為numpy數組需要花費大量時間。 為了實現收支平衡，看起來這些年份的跨度需要大約為30：

開始年/年 - （1985,2013）：

np_fill took 0.049400806427 seconds.
tehsockz_fill took 0.0425939559937 seconds.
zeke_fill_fill took 0.0748357772827 seconds.

Numpy當然會從那里逐漸變得更好。 如果你因任何原因需要返回一個numpy數組，numpy算法總是更快。

Answer 2

minyear = 2008
maxyear = 2013
new_a = []
for group in A:
    group = group
    years = [point[0] for point in group]
    print years
    for year in range(minyear,maxyear+1):
        if year not in years:
            group.append([year,0])
    new_a.append(sorted(group))
print new_a

這會產生：

[   [[2008, 5], [2009, 5], [2010, 2], [2011, 5], [2012, 0], [2013, 17]],
    [[2008, 6], [2009, 3], [2010, 0], [2011, 1], [2012, 0], [2013, 6]],
    [[2008, 0], [2009, 0], [2010, 0], [2011, 0], [2012, 0], [2013, 9]],
    [[2008, 4], [2009, 0], [2010, 0], [2011, 1], [2012, 0], [2013, 4]],
    [[2008, 0], [2009, 0], [2010, 3], [2011, 3], [2012, 0], [2013, 1]],
    [[2008, 2], [2009, 0], [2010, 0], [2011, 4], [2012, 0], [2013, 1]],
    [[2008, 0], [2009, 1], [2010, 1], [2011, 3], [2012, 0], [2013, 3]],
    [[2008, 0], [2009, 0], [2010, 1], [2011, 1], [2012, 0], [2013, 5]],
    [[2008, 0], [2009, 0], [2010, 0], [2011, 1], [2012, 0], [2013, 4]],
    [[2008, 0], [2009, 1], [2010, 0], [2011, 0], [2012, 0], [2013, 4]],
    [[2008, 1], [2009, 0], [2010, 0], [2011, 0], [2012, 0], [2013, 3]],
    [[2008, 0], [2009, 1], [2010, 0], [2011, 0], [2012, 0], [2013, 2]],
    [[2008, 0], [2009, 0], [2010, 0], [2011, 0], [2012, 0], [2013, 2]],
    [[2008, 0], [2009, 0], [2010, 0], [2011, 1], [2012, 0], [2013, 1]],
    [[2008, 0], [2009, 0], [2010, 0], [2011, 0], [2012, 0], [2013, 1]],
    [[2008, 0], [2009, 0], [2010, 0], [2011, 0], [2012, 0], [2013, 1]],
    [[2008, 0], [2009, 0], [2010, 0], [2011, 1], [2012, 0], [2013, 0]],
    [[2008, 0], [2009, 0], [2010, 0], [2011, 1], [2012, 0], [2013, 0]]]

Answer 3

你走了，希望你喜歡它！

min_year = 2007 # for testing purposes I used these years
max_year = 2014

final_list = [] # you're going to be adding to this list the corrected values

for outer in A: # start by iterating through each outer list in A
    active_years = {} # use this dictionary to keep track of which years are in each list and their values; sorry if you don't know about dictionaries

    for inner in outer: # now iterate through each year in each of the outer lists and create a dictionary entry for each (print to see what it's doing)
        active_years[inner[0]] = inner[1] # see who I'm creating a new key-value pair with the key as the year given by the 0th index of inner

    new_outer = [] # this will be your new outer list
    for year in range(min_year, max_year + 1): # now add to your active_years dictionary all the other years and give them value 0
        if year not in active_years.keys(): # only add the years not in your dictionary already
            active_years[year] = 0

    for entry in active_years.keys(): # we now iterate through each key, in order
        new_outer += [[entry, active_years[entry]]] # create your new outer list, watch carefully the brackets
    final_list += [new_outer] # add to the final_list

print final_list # presto