[英]How to multiply selected columns from different pandas dataframes
I have 3 pandas dataframes (similar to the below one). 我有3个pandas数据框(类似于下面的数据框)。 I have 2 lists list ID_1 = ['sdf', 'sdfsdf', ...]
and list ID_2 = ['kjdf', 'kldfjs', ...]
我有2个列表, list ID_1 = ['sdf', 'sdfsdf', ...]
和list ID_2 = ['kjdf', 'kldfjs', ...]
Table1:
ID_1 ID_2 Value
0 PUFPaY9 NdYWqAJ 0.002
1 Iu6AxdB qANhGcw 0.01
2 auESFwW jUEUNdw 0.2345
3 LWbYpca G3uZ_Rg 0.0835
4 8fApIAM mVHrayg 0.0295
Table2:
ID_1 weight1 weight2 .....weightN
0 PUFPaY9
1 Iu6AxdB
2 auESFwW
3 LWbYpca
Table3:
ID_2 weight1 weight2 .....weightN
0 PUFPaY9
1 Iu6AxdB
2 auESFwW
3 LWbYpca
I want to have one dataframe which should be calculated like, 我想有一个应该计算的数据框,
for each x ID_1 in list1:
for each y ID_2 in list2:
if x-y exist in Table1:
temp_row = ( x[weights[i]].* y[weights[i]])
# here i want one to one multiplication, x[weight1]*y[weight1] , x[weight2]*y[weight2]
temp_row.append(value[x-y] in Table1)
new_dataframe.append(temp_row)
return new_dataframe
The required new_dataframe should look like Table4: 所需的new_dataframe应该类似于表4:
Table4:
weight1 weight2 weight3 .....weightN value
0
1
2
3
What I am able to do now is: 我现在能做的是:
new_df = df[(df.ID_1.isin(list1)) & (df.ID_2.isin(list2))]
using this I am getting all valid ID_1
and ID_2
combination and values. new_df = df[(df.ID_1.isin(list1)) & (df.ID_2.isin(list2))]
使用此方法,我将获得所有有效的ID_1
和ID_2
组合和值。 But I have no idea, how I can get the multiplication of weights form both datafames ( without looping for each weight[i]
)? 但是我不知道如何从两个数据帧中获得权重的乘积(而不为每个weight[i]
循环)?
Now task is easier, I can iterate over the new_df
and for each row in new_df
, I will find weight[i to n] for ID_1 from table 2
and weight[i to n] for ID_2 from table3
. 现在任务变得更容易了,我可以遍历new_df
, for each row in new_df
new_df
for each row in new_df
,我weight[i to n] for ID_1 from table 2
找到weight[i to n] for ID_1 from table 2
weight[i to n] for ID_2 from table3
。 Then I can append their one-one multiplication
with "value" from table1
to new FINAL_DF
. 然后,我可以追加其one-one multiplication
与"value" from table1
新FINAL_DF
。 But I don't want to loop and do, can we solve this using some smarter way? 但是我不想循环执行,我们可以使用更智能的方式解决此问题吗?
is that what you want? 那是你要的吗?
data = """\
ID_1
PUFPaY9
aaaaaaa
Iu6AxdB
auESFwW
LWbYpca
"""
id1 = pd.read_csv(io.StringIO(data), delim_whitespace=True)
data = """\
ID_2
PUFPaY9
Iu6AxdB
xxxxxxx
auESFwW
LWbYpca
"""
id2 = pd.read_csv(io.StringIO(data), delim_whitespace=True)
cols = ['weight{}'.format(i) for i in range(1,5)]
for c in cols:
id1[c] = np.random.randint(1, 10, len(id1))
id2[c] = np.random.randint(1, 10, len(id2))
id1.set_index('ID_1', inplace=True)
id2.set_index('ID_2', inplace=True)
df_mul = id1 * id2
Step by step: 一步步:
In [215]: id1
Out[215]:
weight1 weight2 weight3 weight4
ID_1
PUFPaY9 8 9 1 1
aaaaaaa 6 1 9 2
Iu6AxdB 8 4 8 5
auESFwW 9 3 4 2
LWbYpca 7 7 1 8
In [216]: id2
Out[216]:
weight1 weight2 weight3 weight4
ID_2
PUFPaY9 6 5 5 1
Iu6AxdB 1 5 4 5
xxxxxxx 1 2 6 4
auESFwW 3 9 5 5
LWbYpca 3 3 6 7
In [217]: id1 * id2
Out[217]:
weight1 weight2 weight3 weight4
Iu6AxdB 8.0 20.0 32.0 25.0
LWbYpca 21.0 21.0 6.0 56.0
PUFPaY9 48.0 45.0 5.0 1.0
aaaaaaa NaN NaN NaN NaN
auESFwW 27.0 27.0 20.0 10.0
xxxxxxx NaN NaN NaN NaN
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