Python 使用两行时操作数无法一起广播错误

使用以下数据：

import datetime,numpy as np,pandas as pd
nan = np.nan

a = pd.DataFrame(  {'price': {datetime.time(9, 0): 1,   datetime.time(10, 0): 0,   datetime.time(11, 0): 3,   datetime.time(12, 0): 4,   datetime.time(13, 0): 7,   datetime.time(14, 0): 6,   datetime.time(15, 0): 5,   datetime.time(16, 0): 4,   datetime.time(17, 0): 0,   datetime.time(18, 0): 2,   datetime.time(19, 0): 4,   datetime.time(20, 0): 7},  'reversal': {datetime.time(9, 0): 1,   datetime.time(10, 0): nan,   datetime.time(11, 0): nan,   datetime.time(12, 0): nan,   datetime.time(13, 0): nan,
  datetime.time(14, 0): 6.0,   datetime.time(15, 0): nan,   datetime.time(16, 0): nan,   datetime.time(17, 0): nan,   datetime.time(18, 0): nan,   datetime.time(19, 0): nan,   datetime.time(20, 0): nan}})


a['target_hit']=nan;
a['target_miss']=nan;
a['reversal1']=a['reversal']+1;
a['reversal2']=a['reversal']-a['reversal'];
a.sort_index(1,inplace=True);

我创建一个子集，如下所示：

hit = a.ix[:,:-2].dropna()
hit

这将给出显示两行匹配的输出：

         price  reversal  reversal1 reversal2
 09:00:00   1   1.0         2.0     0.0
 14:00:00   6   6.0         7.0     0.0

然后，当我尝试使用这些行来匹配以下内容时，我得到了以下错误

ValueError:操作数无法与形状（2，）（12，）一起广播

takeBoth=False
targetIsHit，targetIsMiss=False，False
如果两者都有：
targetHit=a[（hit['reversal1'].values==a['price'].values）&（hit['reversal1'].index.values是的，它必须用于循环…现在我将其更改为从hit或miss中选择第一个

import datetime,numpy as np,pandas as pd;
nan = np.nan;

a = pd.DataFrame(  {'price': {datetime.time(9, 0): 1,   datetime.time(10, 0): 0,   datetime.time(11, 0): 3,   datetime.time(12, 0): 4,   datetime.time(13, 0): 7,   datetime.time(14, 0): 6,   datetime.time(15, 0): 5,   datetime.time(16, 0): 4,   datetime.time(17, 0): 0,   datetime.time(18, 0): 2,   datetime.time(19, 0): 4,   datetime.time(20, 0): 7},  'reversal': {datetime.time(9, 0): 1,   datetime.time(10, 0): nan,   datetime.time(11, 0): nan,   datetime.time(12, 0): nan,   datetime.time(13, 0): nan,
  datetime.time(14, 0): 6.0,   datetime.time(15, 0): nan,   datetime.time(16, 0): nan,   datetime.time(17, 0): nan,   datetime.time(18, 0): nan,   datetime.time(19, 0): nan,   datetime.time(20, 0): nan}});


a['target_hit']=a['target_miss']=nan;
a['reversal1']=a['reversal']+1;
a['reversal2']=a['reversal']-a['reversal'];
a.sort_index(1,inplace=True);

hits = a.ix[:,:-2].dropna();

for row,hit in hits.iterrows():

        forwardRows = [row]<a['price'].index.values

        targetHit = a.index.values[(hit['reversal1']==a['price'].values) & forwardRows][0];
        targetMiss = a.index.values[(hit['reversal2']==a['price'].values) & forwardRows][0];

        if targetHit>targetMiss:
            a.loc[row,"target_miss"] = targetMiss;
        else:
            a.loc[row,"target_hit"] = targetHit;


print '#'*50
print a
'''

当您执行
hit['reversal1']时会发生错误。值==a['price'].值
。您不应该像这样进行元素级比较（例如使用
np.equals
）。但是，在您的特定情况下，我会确保
hit['reversal1']
的计算结果为标量，例如，我会将
hit
替换为
hit
只选择它的第一行。你能告诉我怎么做@Kris吗？在你的第二个单元格中尝试
hit=hit.iloc[0]
，然后尝试评估你的第三个单元格。再次感谢。这回答了我的问题。如果你有时间看，我还有另一个相关问题
import datetime,numpy as np,pandas as pd;
nan = np.nan;

a = pd.DataFrame(  {'price': {datetime.time(9, 0): 1,   datetime.time(10, 0): 0,   datetime.time(11, 0): 3,   datetime.time(12, 0): 4,   datetime.time(13, 0): 7,   datetime.time(14, 0): 6,   datetime.time(15, 0): 5,   datetime.time(16, 0): 4,   datetime.time(17, 0): 0,   datetime.time(18, 0): 2,   datetime.time(19, 0): 4,   datetime.time(20, 0): 7},  'reversal': {datetime.time(9, 0): 1,   datetime.time(10, 0): nan,   datetime.time(11, 0): nan,   datetime.time(12, 0): nan,   datetime.time(13, 0): nan,
  datetime.time(14, 0): 6.0,   datetime.time(15, 0): nan,   datetime.time(16, 0): nan,   datetime.time(17, 0): nan,   datetime.time(18, 0): nan,   datetime.time(19, 0): nan,   datetime.time(20, 0): nan}});


a['target_hit']=a['target_miss']=nan;
a['reversal1']=a['reversal']+1;
a['reversal2']=a['reversal']-a['reversal'];
a.sort_index(1,inplace=True);

hits = a.ix[:,:-2].dropna();

for row,hit in hits.iterrows():

        forwardRows = [row]<a['price'].index.values

        targetHit = a.index.values[(hit['reversal1']==a['price'].values) & forwardRows][0];
        targetMiss = a.index.values[(hit['reversal2']==a['price'].values) & forwardRows][0];

        if targetHit>targetMiss:
            a.loc[row,"target_miss"] = targetMiss;
        else:
            a.loc[row,"target_hit"] = targetHit;


print '#'*50
print a
'''

    ##################################################
##################################################
          price  reversal  reversal1  reversal2  target_hit target_miss
09:00:00      1       1.0        2.0        0.0         NaN    10:00:00
10:00:00      0       NaN        NaN        NaN         NaN         NaN
11:00:00      3       NaN        NaN        NaN         NaN         NaN
12:00:00      4       NaN        NaN        NaN         NaN         NaN
13:00:00      7       NaN        NaN        NaN         NaN         NaN
14:00:00      6       6.0        7.0        0.0         NaN    17:00:00
15:00:00      5       NaN        NaN        NaN         NaN         NaN
16:00:00      4       NaN        NaN        NaN         NaN         NaN
17:00:00      0       NaN        NaN        NaN         NaN         NaN
18:00:00      2       NaN        NaN        NaN         NaN         NaN
19:00:00      4       NaN        NaN        NaN         NaN         NaN
20:00:00      7       NaN        NaN        NaN         NaN         NaN

'''