Question

我在下面有以下Python数据框。＆＃34; Flag＆＃34;字段是我想用代码创建的所需列。

我想做以下事情：

如果＆＃34;分配类型＆＃34;是预测和＆＃34; Activities_Counter＆＃34;大于10，我想创建一个名为＆＃34; Flag＆＃34;的新列。并使用＆＃39;标记＆＃39;标记该行。

否则，将Flag行留空。

我使用以下代码来识别/标记＆＃34; Activities_Counter＆＃34;大于10 ...但我不知道如何纳入＆＃34;分配类型＆＃34;标准到我的代码中。

Flag = []

for row in df_HA_noHA_act['Activities_Counter']:
    if row >= 10:
        Flag.append('Flag')
    else:
        Flag.append('')

df_HA_noHA_act['Flag'] = Flag

非常感谢任何帮助！

Answer 1

您需要使用&添加新条件。使用numpy.where：

也更快

mask = (df_HA_noHA_act["Allocation Type"] == 'Predicted') & 
       (df_HA_noHA_act['Activities_Counter'] >= 10)
df_HA_noHA_act['Flag'] = np.where(mask, 'Flag', '')

df_HA_noHA_act = pd.DataFrame({'Activities_Counter':[10,2,6,15,11,18],
                              'Allocation Type':['Historical','Historical','Predicted', 
                                                  'Predicted','Predicted','Historical']})
print (df_HA_noHA_act)
   Activities_Counter Allocation Type
0                  10      Historical
1                   2      Historical
2                   6       Predicted
3                  15       Predicted
4                  11       Predicted
5                  18      Historical

mask = (df_HA_noHA_act["Allocation Type"] == 'Predicted') & 
       (df_HA_noHA_act['Activities_Counter'] >= 10)
df_HA_noHA_act['Flag'] = np.where(mask, 'Flag', '')
print (df_HA_noHA_act)
   Activities_Counter Allocation Type  Flag
0                  10      Historical      
1                   2      Historical      
2                   6       Predicted      
3                  15       Predicted  Flag
4                  11       Predicted  Flag
5                  18      Historical

循环缓慢解决方案：

Flag = []
for i, row in df_HA_noHA_act.iterrows():
    if (row['Activities_Counter'] >= 10) and (row["Allocation Type"] == 'Predicted'):
        Flag.append('Flag')
    else:
        Flag.append('')
df_HA_noHA_act['Flag'] = Flag
print (df_HA_noHA_act)
   Activities_Counter Allocation Type  Flag
0                  10      Historical      
1                   2      Historical      
2                   6       Predicted      
3                  15       Predicted  Flag
4                  11       Predicted  Flag
5                  18      Historical

<强>计时：

df_HA_noHA_act = pd.DataFrame({'Activities_Counter':[10,2,6,15,11,18],
                              'Allocation Type':['Historical','Historical','Predicted', 
                                                  'Predicted','Predicted','Historical']})
print (df_HA_noHA_act)
#[6000 rows x 2 columns]
df_HA_noHA_act = pd.concat([df_HA_noHA_act]*1000).reset_index(drop=True)

In [187]: %%timeit
     ...: df_HA_noHA_act['Flag1'] = np.where((df_HA_noHA_act["Allocation Type"] == 'Predicted') & (df_HA_noHA_act['Activities_Counter'] >= 10), 'Flag', '')
     ...: 
100 loops, best of 3: 1.89 ms per loop

In [188]: %%timeit
     ...: Flag = []
     ...: for i, row in df_HA_noHA_act.iterrows():
     ...:     if (row['Activities_Counter'] >= 10) and (row["Allocation Type"] == 'Predicted'):
     ...:         Flag.append('Flag')
     ...:     else:
     ...:         Flag.append('')
     ...: df_HA_noHA_act['Flag'] = Flag
     ...: 
     ...: 
1 loop, best of 3: 381 ms per loop

Python数据框：根据字符串列和浮点列中的值创建新列

1 个答案: