內插數字序列

Question

我正在嘗試填寫不完整的數字列表，找不到任何Python方式來實現。 我有一系列從1到31的天，並且每天都有一個float值。

#dictionnary{day: value}
monthvalues = {1: 1.12, 2: 3.24, 3: 2.23, 5: 2.10, 7: 4.97} etc.. to 31st day

但是我的數據不完整，有些日子還沒到！ 因此，我想用這種方式在數學上填補丟失的圖片：

樣本月份1：

{16: 2.00, 18: 4.00}
#==> I want to add to the dictionnary 17: 3.00

樣本月份2：

{10: 2.00, 14: 4.00}
#==> I want to add to the dictionnary 11: 2.25, 12: 2.50, 13: 2.75

聽起來很簡單，但是我有數百萬行要從一個不完整的sql數據庫中處理，並且目前我對xrange（）循環迷失了……也許在數學庫中有一個方法，但我找不到它。

謝謝你的幫助！

編輯：我想對數字進行插值，但據我所知，只有numpy / scipy具有此類數學函數，並且im使用與numpy / scipy不兼容的Pypy。

Answer 1

考慮為此使用pandas ， interpolate方法很容易：

In [502]: import pandas    

In [503]: s = pandas.Series({1: 1.12, 2: 3.24, 3: 2.23,5: 2.10,7:4.97}, index=range(1,8))

In [504]: s
Out[504]: 
1    1.12
2    3.24
3    2.23
4     NaN
5    2.10
6     NaN
7    4.97

In [505]: s.interpolate()
Out[505]: 
1    1.120
2    3.240
3    2.230
4    2.165
5    2.100
6    3.535
7    4.970

並且具有多個缺失值：

In [506]: s2 = pandas.Series({10: 2.00, 14: 4.00},index=range(10,15))

In [507]: s2
Out[507]: 
10     2
11   NaN
12   NaN
13   NaN
14     4

In [508]: s2.interpolate()
Out[508]: 
10    2.0
11    2.5
12    3.0
13    3.5
14    4.0

如果需要，您可以將其轉換回字典。

In [511]: s2.to_dict()
Out[511]: {10: 2.0, 11: 2.5, 12: 3.0, 13: 3.5, 14: 4.0}

Answer 2

您只需要一些簡單的循環和良好的舊編程邏輯即可。 此邏輯的一個警告是，您需要一個開始和結束編號才能使其正常工作。 我不知道這對您的數據是否有意義，但是插值法要求這樣做。

設定：

# Keeps track of the last "seen" day
lastday=0

# Default 1st day if missing
if 1 not in monthvalues:
  monthvalues[1] = 1.23 #you need a default

# Default 31st day if missing
if 31 not in monthvalues:
  monthvalues[31] = 1.23 #you need a default

處理：

# Loop from 1 to 31
for thisday in range(1,32):

  # If we do not encounter thisday in the monthvalues, then skip and keep looping
  if thisday not in monthvalues:
    continue

  # How far ago was the last day seen?
  gap = thisday - lastday

  # If the last day was more than 1 ago, it means there is at least one day amis
  if gap > 1:

    # This is the amount of the last "seen" day
    last_amt = monthvalues[lastday]

    # this is the difference between the current day and the last day
    diff = monthvalues[thisday] - last_amt

    # This is how much you want to interpolate per day in-between
    amt_per_day = diff/gap

    # there is a gap of missing days, let's fill them
    # Start at 1 because we start at the day after the last seen day
    for n in range(1, gap):

      # Fill the missing days with an interpolated value
      monthvalues[lastday+n] = last_amt + amt_per_day * n

  # For the next iteration of the loop, this is the last seen day.
  lastday = thisday

Answer 3

我認為使用scipy的插值方法是一種聰明的方法

首先將您的數據轉換為易於操作的格式：

monthvalue = {1: 1.12, 2: 3.24, 3: 2.23, 5: 2.10, 7: 4.97, 6: 3.10, 10: 3.3}
X = sorted(monthvalue.keys())
Y = [monthvalue[x] for x in X]

然后創建線性插值函數並輸出中間值

# interpolate function
f = interp1d(X, Y, kind='linear')

x_new = range(X[0], X[-1]+1, 1)
for x in x_new:
    print "%s: %s" % (x, f(x))

結果：

1: 1.12
2: 3.24
3: 2.23
4: 2.165
5: 2.1
6: 3.1
7: 4.97
8: 4.41333333333
9: 3.85666666667
10: 3.3