填写一个列表/熊猫.包含所有缺失数据组合的数据框架(如R中的complete())

Question

我有如下数据集(这是一个示例，它实际上有66k行):

        Type       Food      Loc  Num
0      Fruit     Banana  House-1   15
1      Fruit     Banana  House-2    4
2      Fruit      Apple  House-2    6
3      Fruit      Apple  House-3    8
4  Vegetable   Broccoli  House-3    8
5  Vegetable    Lettuce  House-4   12
6  Vegetable    Peppers  House-5    3
7  Vegetable       Corn  House-4    4
8  Seasoning  Olive Oil  House-6    2
9  Seasoning    Vinegar  House-7    2

我想填满所有缺失的组合(房子3-7有多少香蕉?)例如，除了House-5之外，还有多少个辣椒?)输入0，得到如下结果:

        Type       Food      Loc  Num
0      Fruit     Banana  House-1   15
1      Fruit     Banana  House-2    4
2      Fruit     Banana  House-3    0
... fill remaining houses with zeros
6      Fruit     Banana  House-7    0
7      Fruit      Apple  House-1    0
8      Fruit      Apple  House-2    6
9      Fruit      Apple  House-3    8
... fill remaining houses with zeros
14  Vegetable   Broccoli  House-1    0
15  Vegetable   Broccoli  House-2    0
16  Vegetable   Broccoli  House-3    8
... etc    
n   Seasoning    Vinegar  House-7    2

我知道R有complete函数的积分

现在我正在处理从原始DataFrame中提取的列表，我将其转换为字典。

for key,grp in fruit.groupby(level=0):
        dir[key] = test.ix[key].values.tolist()
fruit = {'Banana': [[1.0,15.0], [2.0,4.0],
         'Apple': [[2.0,6.0], [3.0,8.0]
#Type = {fruit1:[[Loc1,Count1],...,[Locn],[Countn],
#... fruitn:[...]}

我设计了这个函数来应用于字典的赋值规则:

def fill_zeros(list):
    final = [0] * 127
    for i in list:
        final[int(i[0])] = i[1]
    return final

适用于个别的"水果":

print fill_zeros(test.ix['QLLSEEEKK'].values.tolist())
print fill_zeros(test.ix['GAVPLEMLEIALR'].values.tolist())
print fill_zeros(test.ix['VPVNLLNSPDCDVK'].values.tolist())

但是不在字典上:

for key,grp in test.groupby(level=0):
        dir[key] = fill_zeros(test.ix[key].values.tolist())
Traceback (most recent call last):
  File "peptidecount.py", line 59, in <module>
    print fill_zeros(test.ix[str(key)].values.tolist())
  File "peptidecount.py", line 43, in fill_zeros
    final[int(i[0])] = i[1]
TypeError: 'float' object has no attribute '__getitem__'

显然我没有正确地在字典上迭代。有办法纠正吗?或者是否有更合适的函数直接应用于DataFrame?

Answer 1

您可以使用reindex

首先你需要一个有效的(type, food)对的列表。我将从数据本身得到它，而不是把它们写出来。

In [88]: kinds = list(df[['Type', 'Food']].drop_duplicates().itertuples(index=False))
In [89]: kinds
Out[89]:
[('Fruit', 'Banana'),
 ('Fruit', 'Apple'),
 ('Vegetable', 'Broccoli'),
 ('Vegetable', 'Lettuce'),
 ('Vegetable', 'Peppers'),
 ('Vegetable', 'Corn'),
 ('Seasoning', 'Olive Oil'),
 ('Seasoning', 'Vinegar')]

现在，我们将生成kinds与使用itertools.product的房屋的所有配对。

In [93]: from itertools import product
In [94]: houses = ['House-%s' % x for x in range(1, 8)]
In [95]: idx = [(x.Type, x.Food, house) for x, house in product(kinds, houses)]
In [96]: idx[:2]
Out[96]: [('Fruit', 'Banana', 'House-1'), ('Fruit', 'Banana', 'House-2')]

现在你可以使用set_index和reindex来获得缺失的观测值。

In [98]: df.set_index(['Type', 'Food', 'Loc']).reindex(idx, fill_value=0)
Out[98]:
                           Num
Type      Food    Loc
Fruit     Banana  House-1   15
                  House-2    4
                  House-3    0
                  House-4    0
                  House-5    0
...                        ...
Seasoning Vinegar House-3    0
                  House-4    0
                  House-5    0
                  House-6    0
                  House-7    2
[56 rows x 1 columns]

Answer 2

应该可以:

cond0 = df.Num.isnull()
cond1 = df.Food == 'Banana'
cond2 = df.Loc.str.match(r'House-[34567]')
cond3 = df.Food == 'Peppers'
cond4 = df.Loc != 'House-5'
missing_bananas = cond0 & cond1 & cond2
missing_peppers = cond0 & cond3 & cond4
missing_food = missing_bananas | missing_peppers
df.loc[missing_food] = df.loc[missing_food].fillna(0)