如何选择不只包含NaN值和0的行

Question

这是我的数据帧：

cols = ['Country', 'Year', 'Orange', 'Apple', 'Plump']

data = [['US', 2008, 17, 29, 19],
        ['US', 2009, 11, 12, 16],
        ['US', 2010, 14, 16, 38],
        ['Spain', 2008, 11, None, 33],
        ['Spain', 2009, 12, 19, 17],
        ['France', 2008, 17, 19, 21],
        ['France', 2009, 19, 22, 13],
        ['France', 2010, 12, 11, 0],
        ['France', 2010, 0, 0, 0],
        ['Italy', 2009, None, None, None],
        ['Italy', 2010, 15, 16, 17],
        ['Italy', 2010, 0, None, None],
        ['Italy', 2011, 42, None, None]]

我要选择的行中，橙色苹果和丰满不只是由“无”，只有0或它们的混合。所以结果应该是：

   Country  Year  Orange  Apple  Plump  
0       US  2008    17.0   29.0   19.0  
1       US  2009    11.0   12.0   16.0  
2       US  2010    14.0   16.0   38.0  
3    Spain  2008    11.0    NaN   33.0  
4    Spain  2009    12.0   19.0   17.0  
5   France  2008    17.0   19.0   21.0 
6   France  2009    19.0   22.0   13.0  
7   France  2010    12.0   11.0    0.0  
10   Italy  2010    15.0   16.0   17.0  
12   Italy  2011    42.0    NaN    NaN  

第二，我想放弃那些我三年都没有观察到的国家。因此，由此产生的产出应该只有美国和法国。我怎样才能得到它们？ 我试过这样的方法：

df = df[(df['Orange'].notnull())| \
            (df['Apple'].notnull()) | (df['Plump'].notnull()) | (df['Orange'] != 0 )| (df['Apple']!= 0) | (df['Plump']!= 0)]

我也试过：

df = df[((df['Orange'].notnull())| \
                (df['Apple'].notnull()) | (df['Plump'].notnull())) & ((df['Orange'] != 0 )| (df['Apple']!= 0) | (df['Plump']!= 0))]

Answer 1

没有值将被读取为NaN，因此您可以替换0并将它们转换为NaN。在那之后你就可以按照马苏的建议去做了。可能是这样的：

In: df = df.replace(0,np.nan)
    df = df[df[['Orange','Apple','Plump']].notnull().any(1)]
Out:
   Country  Year  Orange  Apple  Plump

0       US  2008      17     29     19
1       US  2009      11     12     16
2       US  2010      14     16     38
3    Spain  2008      11    NaN     33
4    Spain  2009      12     19     17
5   France  2008      17     19     21
6   France  2009      19     22     13
7   France  2010      12     11    NaN
10   Italy  2010      15     16     17
12   Italy  2011      42    NaN    NaN

对于你的第二个问题，我理解在这个例子中，你想摆脱那些你没有200820092010年观察数据的国家。 为此，你可以采取如下措施：

countries = []
for group,values in enumerate(df.groupby('Country')):
    lista = values[1].Year.unique() == [2008,2009,2010]
    if (np.all(lista)):
        countries.append(values[0])
df = df[df.Country.isin(countries)]

这将产生如下结果：

  Country  Year  Orange  Apple  Plump
0      US  2008      17     29     19
1      US  2009      11     12     16
2      US  2010      14     16     38
5  France  2008      17     19     21
6  France  2009      19     22     13
7  France  2010      12     11    NaN
8  France  2010     NaN    NaN    NaN    

最后，您可以同时应用这两种解决方案：

df[df[['Orange','Apple','Plump']].notnull().any(1) & df.Country.isin(countries)])

获取：

  Country  Year  Orange  Apple  Plump
0      US  2008      17     29     19
1      US  2009      11     12     16
2      US  2010      14     16     38
5  France  2008      17     19     21
6  France  2009      19     22     13   
7  France  2010      12     11    NaN

Answer 2

In [307]: df[~df[['Orange','Apple','Plump']].fillna(0).eq(0).all(1)]
Out[307]:
   Country  Year  Orange  Apple  Plump
0       US  2008    17.0   29.0   19.0
1       US  2009    11.0   12.0   16.0
2       US  2010    14.0   16.0   38.0
3    Spain  2008    11.0    NaN   33.0
4    Spain  2009    12.0   19.0   17.0
5   France  2008    17.0   19.0   21.0
6   France  2009    19.0   22.0   13.0
7   France  2010    12.0   11.0    0.0
10   Italy  2010    15.0   16.0   17.0
12   Italy  2011    42.0    NaN    NaN