基于条件选择的新列，来自Pandas DataFrame中其他2列的值

Question

I've got a DataFrame which contains stock values.

It looks like this:

>>>Data Open High Low Close Volume Adj Close Date                                                       
2013-07-08  76.91  77.81  76.85  77.04  5106200  77.04

When I try to make a conditional new column with the following if statement:

Data['Test'] =Data['Close'] if Data['Close'] > Data['Open'] else Data['Open']

I get the following error:

Traceback (most recent call last):
  File "<pyshell#116>", line 1, in <module>
    Data[1]['Test'] =Data[1]['Close'] if Data[1]['Close'] > Data[1]['Open'] else Data[1]['Open']
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

I then used a.all() :

Data[1]['Test'] =Data[1]['Close'] if all(Data[1]['Close'] > Data[1]['Open']) else Data[1]['Open']

The result was that the entire ['Open'] Column was selected. I didn't get the condition that I wanted, which is to select every time the biggest value between the ['Open'] and ['Close'] columns.

Any help is appreciated.

Thanks.

Answer 1

From a DataFrame like:

>>> df
         Date   Open   High    Low  Close   Volume  Adj Close
0  2013-07-08  76.91  77.81  76.85  77.04  5106200      77.04
1  2013-07-00  77.04  79.81  71.81  72.87  1920834      77.04
2  2013-07-10  72.87  99.81  64.23  93.23  2934843      77.04

The simplest thing I can think of would be:

>>> df["Test"] = df[["Open", "Close"]].max(axis=1)
>>> df
         Date   Open   High    Low  Close   Volume  Adj Close   Test
0  2013-07-08  76.91  77.81  76.85  77.04  5106200      77.04  77.04
1  2013-07-00  77.04  79.81  71.81  72.87  1920834      77.04  77.04
2  2013-07-10  72.87  99.81  64.23  93.23  2934843      77.04  93.23

df.ix[:,["Open", "Close"]].max(axis=1) might be a little faster, but I don't think it's as nice to look at.

Alternatively, you could use .apply on the rows:

>>> df["Test"] = df.apply(lambda row: max(row["Open"], row["Close"]), axis=1)
>>> df
         Date   Open   High    Low  Close   Volume  Adj Close   Test
0  2013-07-08  76.91  77.81  76.85  77.04  5106200      77.04  77.04
1  2013-07-00  77.04  79.81  71.81  72.87  1920834      77.04  77.04
2  2013-07-10  72.87  99.81  64.23  93.23  2934843      77.04  93.23

Or fall back to numpy:

>>> df["Test"] = np.maximum(df["Open"], df["Close"])
>>> df
         Date   Open   High    Low  Close   Volume  Adj Close   Test
0  2013-07-08  76.91  77.81  76.85  77.04  5106200      77.04  77.04
1  2013-07-00  77.04  79.81  71.81  72.87  1920834      77.04  77.04
2  2013-07-10  72.87  99.81  64.23  93.23  2934843      77.04  93.23

The basic problem is that if/else doesn't play nicely with arrays, because if (something) always coerces the something into a single bool . It's not equivalent to "for every element in the array something, if the condition holds" or anything like that.

Answer 2

In [7]: df = DataFrame(randn(10,2),columns=list('AB'))

In [8]: df
Out[8]: 
          A         B
0 -0.954317 -0.485977
1  0.364845 -0.193453
2  0.020029 -1.839100
3  0.778569  0.706864
4  0.033878  0.437513
5  0.362016  0.171303
6  2.880953  0.856434
7 -0.109541  0.624493
8  1.015952  0.395829
9 -0.337494  1.843267

This is a where conditional, saying give me the value for A if A > B, else give me B

# this syntax is EQUIVALENT to
# df.loc[df['A']>df['B'],'A'] = df['B']

In [9]: df['A'].where(df['A']>df['B'],df['B'])
Out[9]: 
0   -0.485977
1    0.364845
2    0.020029
3    0.778569
4    0.437513
5    0.362016
6    2.880953
7    0.624493
8    1.015952
9    1.843267
dtype: float64

In this case max is equivalent

In [10]: df.max(1)
Out[10]: 
0   -0.485977
1    0.364845
2    0.020029
3    0.778569
4    0.437513
5    0.362016
6    2.880953
7    0.624493
8    1.015952
9    1.843267
dtype: float64

Answer 3

The issue is that you're asking python to evaluate a condition ( Data['Close'] > Data['Open'] ) which contains more than one boolean value. You do not want to use any or all since either, since that will set Data['Test'] to either Data['Open'] or Data['Close'] .

There might be a cleaner method, but one approach is to use a mask (boolean array):

mask = Data['Close'] > Data['Open']
Data['Test'] = pandas.concat([Data['Close'][mask].dropna(), Data['Open'][~mask].dropna()]).reindex_like(Data)