applying row-wise function in a data frame with values of another data frame
0
I have two data frames:
df1
df1 = pd.DataFrame( 'group' : ["A","A","A","B","B","B"],
'par' : [5,5,15,10,2,11],
'val' :[50,10,180,10,10,660],
'set_0' :["Country","Country","Country","Country","Country","Country"],
'set_1' :["size1","size1","size2","size3","size4","size3"],
'set_2' :["size12","size12","size12","size9","size13","size13"],
'set_3' :["size14","size14","size15","NO","NO","NO"],
'set_4' :["NO","NO","NO","size25","size25","size27"],
'set_5' :["NO","NO","NO","NO","NO","NO"]
)
df2
df2 = pd.DataFrame( 'group' : ["A","A","A","A","A","A","B","B","B","B","B","B","B"],
'set' : ["Country","size1","size2","size12","size14","size15","Country","size3","size4","size9","size13","size25","size27"],
)
For each row (group X set combination) of df2, I want to apply a function to calculate (sum of "val"/sum of "par").
I tried to use something with apply function but I could not really figure out since I am pretty new in python.
Could anyone please help regarding a solution?
below is the expected outcome:
As image upload keeps failing, I am also sharing below the code to get the outcome in the ugliest and most inefficient hardcoding way:
a1=df1[(df1["group"]==df2.iloc[0,0])&(df1["set_0"]==df2.iloc[0,1])].sum()["val"]
a2=df1[(df1["group"]==df2.iloc[0,0])&(df1["set_0"]==df2.iloc[0,1])].sum()["par"]
b1=df1[(df1["group"]==df2.iloc[1,0])&(df1["set_1"]==df2.iloc[1,1])].sum()["val"]
b2=df1[(df1["group"]==df2.iloc[1,0])&(df1["set_1"]==df2.iloc[1,1])].sum()["par"]
c1=df1[(df1["group"]==df2.iloc[2,0])&(df1["set_1"]==df2.iloc[2,1])].sum()["val"]
c2=df1[(df1["group"]==df2.iloc[2,0])&(df1["set_1"]==df2.iloc[2,1])].sum()["par"]
d1=df1[(df1["group"]==df2.iloc[3,0])&(df1["set_2"]==df2.iloc[3,1])].sum()["val"]
d2=df1[(df1["group"]==df2.iloc[3,0])&(df1["set_2"]==df2.iloc[3,1])].sum()["par"]
e1=df1[(df1["group"]==df2.iloc[4,0])&(df1["set_3"]==df2.iloc[4,1])].sum()["val"]
e2=df1[(df1["group"]==df2.iloc[4,0])&(df1["set_3"]==df2.iloc[4,1])].sum()["par"]
f1=df1[(df1["group"]==df2.iloc[5,0])&(df1["set_3"]==df2.iloc[5,1])].sum()["val"]
f2=df1[(df1["group"]==df2.iloc[5,0])&(df1["set_3"]==df2.iloc[5,1])].sum()["par"]
g1=df1[(df1["group"]==df2.iloc[6,0])&(df1["set_0"]==df2.iloc[6,1])].sum()["val"]
g2=df1[(df1["group"]==df2.iloc[6,0])&(df1["set_0"]==df2.iloc[6,1])].sum()["par"]
h1=df1[(df1["group"]==df2.iloc[7,0])&(df1["set_1"]==df2.iloc[7,1])].sum()["val"]
h2=df1[(df1["group"]==df2.iloc[7,0])&(df1["set_1"]==df2.iloc[7,1])].sum()["par"]
j1=df1[(df1["group"]==df2.iloc[8,0])&(df1["set_1"]==df2.iloc[8,1])].sum()["val"]
j2=df1[(df1["group"]==df2.iloc[8,0])&(df1["set_1"]==df2.iloc[8,1])].sum()["par"]
k1=df1[(df1["group"]==df2.iloc[9,0])&(df1["set_2"]==df2.iloc[9,1])].sum()["val"]
k2=df1[(df1["group"]==df2.iloc[9,0])&(df1["set_2"]==df2.iloc[9,1])].sum()["par"]
l1=df1[(df1["group"]==df2.iloc[10,0])&(df1["set_2"]==df2.iloc[10,1])].sum()["val"]
l2=df1[(df1["group"]==df2.iloc[10,0])&(df1["set_2"]==df2.iloc[10,1])].sum()["par"]
m1=df1[(df1["group"]==df2.iloc[11,0])&(df1["set_4"]==df2.iloc[11,1])].sum()["val"]
m2=df1[(df1["group"]==df2.iloc[11,0])&(df1["set_4"]==df2.iloc[11,1])].sum()["par"]
n1=df1[(df1["group"]==df2.iloc[12,0])&(df1["set_4"]==df2.iloc[12,1])].sum()["val"]
n2=df1[(df1["group"]==df2.iloc[12,0])&(df1["set_4"]==df2.iloc[12,1])].sum()["par"]
Up to this part, I had to manually create "val" and "par" pairs for each element.
Then,
a=a1/a2
b=b1/b2
c=c1/c2
d=d1/d2
e=e1/e2
f=f1/f2
g=g1/g2
h=h1/h2
j=j1/j2
k=k1/k2
l=l1/l2
m=m1/m2
n=n1/n2
Finally, the result is:
df2["desired_calculation"]=[a,b,c,d,e,f,g,h,j,k,l,m,n]
python-3.x function apply
asked Nov 13 '18 at 11:48
nkltkfnkltkf
113
add a comment |
0
I have two data frames:
df1
df1 = pd.DataFrame( 'group' : ["A","A","A","B","B","B"],
'par' : [5,5,15,10,2,11],
'val' :[50,10,180,10,10,660],
'set_0' :["Country","Country","Country","Country","Country","Country"],
'set_1' :["size1","size1","size2","size3","size4","size3"],
'set_2' :["size12","size12","size12","size9","size13","size13"],
'set_3' :["size14","size14","size15","NO","NO","NO"],
'set_4' :["NO","NO","NO","size25","size25","size27"],
'set_5' :["NO","NO","NO","NO","NO","NO"]
)
df2
df2 = pd.DataFrame( 'group' : ["A","A","A","A","A","A","B","B","B","B","B","B","B"],
'set' : ["Country","size1","size2","size12","size14","size15","Country","size3","size4","size9","size13","size25","size27"],
)
For each row (group X set combination) of df2, I want to apply a function to calculate (sum of "val"/sum of "par").
I tried to use something with apply function but I could not really figure out since I am pretty new in python.
Could anyone please help regarding a solution?
below is the expected outcome:
As image upload keeps failing, I am also sharing below the code to get the outcome in the ugliest and most inefficient hardcoding way:
a1=df1[(df1["group"]==df2.iloc[0,0])&(df1["set_0"]==df2.iloc[0,1])].sum()["val"]
a2=df1[(df1["group"]==df2.iloc[0,0])&(df1["set_0"]==df2.iloc[0,1])].sum()["par"]
b1=df1[(df1["group"]==df2.iloc[1,0])&(df1["set_1"]==df2.iloc[1,1])].sum()["val"]
b2=df1[(df1["group"]==df2.iloc[1,0])&(df1["set_1"]==df2.iloc[1,1])].sum()["par"]
c1=df1[(df1["group"]==df2.iloc[2,0])&(df1["set_1"]==df2.iloc[2,1])].sum()["val"]
c2=df1[(df1["group"]==df2.iloc[2,0])&(df1["set_1"]==df2.iloc[2,1])].sum()["par"]
d1=df1[(df1["group"]==df2.iloc[3,0])&(df1["set_2"]==df2.iloc[3,1])].sum()["val"]
d2=df1[(df1["group"]==df2.iloc[3,0])&(df1["set_2"]==df2.iloc[3,1])].sum()["par"]
e1=df1[(df1["group"]==df2.iloc[4,0])&(df1["set_3"]==df2.iloc[4,1])].sum()["val"]
e2=df1[(df1["group"]==df2.iloc[4,0])&(df1["set_3"]==df2.iloc[4,1])].sum()["par"]
f1=df1[(df1["group"]==df2.iloc[5,0])&(df1["set_3"]==df2.iloc[5,1])].sum()["val"]
f2=df1[(df1["group"]==df2.iloc[5,0])&(df1["set_3"]==df2.iloc[5,1])].sum()["par"]
g1=df1[(df1["group"]==df2.iloc[6,0])&(df1["set_0"]==df2.iloc[6,1])].sum()["val"]
g2=df1[(df1["group"]==df2.iloc[6,0])&(df1["set_0"]==df2.iloc[6,1])].sum()["par"]
h1=df1[(df1["group"]==df2.iloc[7,0])&(df1["set_1"]==df2.iloc[7,1])].sum()["val"]
h2=df1[(df1["group"]==df2.iloc[7,0])&(df1["set_1"]==df2.iloc[7,1])].sum()["par"]
j1=df1[(df1["group"]==df2.iloc[8,0])&(df1["set_1"]==df2.iloc[8,1])].sum()["val"]
j2=df1[(df1["group"]==df2.iloc[8,0])&(df1["set_1"]==df2.iloc[8,1])].sum()["par"]
k1=df1[(df1["group"]==df2.iloc[9,0])&(df1["set_2"]==df2.iloc[9,1])].sum()["val"]
k2=df1[(df1["group"]==df2.iloc[9,0])&(df1["set_2"]==df2.iloc[9,1])].sum()["par"]
l1=df1[(df1["group"]==df2.iloc[10,0])&(df1["set_2"]==df2.iloc[10,1])].sum()["val"]
l2=df1[(df1["group"]==df2.iloc[10,0])&(df1["set_2"]==df2.iloc[10,1])].sum()["par"]
m1=df1[(df1["group"]==df2.iloc[11,0])&(df1["set_4"]==df2.iloc[11,1])].sum()["val"]
m2=df1[(df1["group"]==df2.iloc[11,0])&(df1["set_4"]==df2.iloc[11,1])].sum()["par"]
n1=df1[(df1["group"]==df2.iloc[12,0])&(df1["set_4"]==df2.iloc[12,1])].sum()["val"]
n2=df1[(df1["group"]==df2.iloc[12,0])&(df1["set_4"]==df2.iloc[12,1])].sum()["par"]
Up to this part, I had to manually create "val" and "par" pairs for each element.
Then,
a=a1/a2
b=b1/b2
c=c1/c2
d=d1/d2
e=e1/e2
f=f1/f2
g=g1/g2
h=h1/h2
j=j1/j2
k=k1/k2
l=l1/l2
m=m1/m2
n=n1/n2
Finally, the result is:
df2["desired_calculation"]=[a,b,c,d,e,f,g,h,j,k,l,m,n]
python-3.x function apply
asked Nov 13 '18 at 11:48
nkltkfnkltkf
113
Can you show us your desired output?
– jpp
Nov 13 '18 at 14:23
@jpp, not sure if the pasted output is gonna look readable, sharing it with an edit.
– nkltkf
Nov 13 '18 at 19:58
add a comment |
0
0
0
I have two data frames:
df1
df1 = pd.DataFrame( 'group' : ["A","A","A","B","B","B"],
'par' : [5,5,15,10,2,11],
'val' :[50,10,180,10,10,660],
'set_0' :["Country","Country","Country","Country","Country","Country"],
'set_1' :["size1","size1","size2","size3","size4","size3"],
'set_2' :["size12","size12","size12","size9","size13","size13"],
'set_3' :["size14","size14","size15","NO","NO","NO"],
'set_4' :["NO","NO","NO","size25","size25","size27"],
'set_5' :["NO","NO","NO","NO","NO","NO"]
)
df2
df2 = pd.DataFrame( 'group' : ["A","A","A","A","A","A","B","B","B","B","B","B","B"],
'set' : ["Country","size1","size2","size12","size14","size15","Country","size3","size4","size9","size13","size25","size27"],
)
For each row (group X set combination) of df2, I want to apply a function to calculate (sum of "val"/sum of "par").
I tried to use something with apply function but I could not really figure out since I am pretty new in python.
Could anyone please help regarding a solution?
below is the expected outcome:
As image upload keeps failing, I am also sharing below the code to get the outcome in the ugliest and most inefficient hardcoding way:
a1=df1[(df1["group"]==df2.iloc[0,0])&(df1["set_0"]==df2.iloc[0,1])].sum()["val"]
a2=df1[(df1["group"]==df2.iloc[0,0])&(df1["set_0"]==df2.iloc[0,1])].sum()["par"]
b1=df1[(df1["group"]==df2.iloc[1,0])&(df1["set_1"]==df2.iloc[1,1])].sum()["val"]
b2=df1[(df1["group"]==df2.iloc[1,0])&(df1["set_1"]==df2.iloc[1,1])].sum()["par"]
c1=df1[(df1["group"]==df2.iloc[2,0])&(df1["set_1"]==df2.iloc[2,1])].sum()["val"]
c2=df1[(df1["group"]==df2.iloc[2,0])&(df1["set_1"]==df2.iloc[2,1])].sum()["par"]
d1=df1[(df1["group"]==df2.iloc[3,0])&(df1["set_2"]==df2.iloc[3,1])].sum()["val"]
d2=df1[(df1["group"]==df2.iloc[3,0])&(df1["set_2"]==df2.iloc[3,1])].sum()["par"]
e1=df1[(df1["group"]==df2.iloc[4,0])&(df1["set_3"]==df2.iloc[4,1])].sum()["val"]
e2=df1[(df1["group"]==df2.iloc[4,0])&(df1["set_3"]==df2.iloc[4,1])].sum()["par"]
f1=df1[(df1["group"]==df2.iloc[5,0])&(df1["set_3"]==df2.iloc[5,1])].sum()["val"]
f2=df1[(df1["group"]==df2.iloc[5,0])&(df1["set_3"]==df2.iloc[5,1])].sum()["par"]
g1=df1[(df1["group"]==df2.iloc[6,0])&(df1["set_0"]==df2.iloc[6,1])].sum()["val"]
g2=df1[(df1["group"]==df2.iloc[6,0])&(df1["set_0"]==df2.iloc[6,1])].sum()["par"]
h1=df1[(df1["group"]==df2.iloc[7,0])&(df1["set_1"]==df2.iloc[7,1])].sum()["val"]
h2=df1[(df1["group"]==df2.iloc[7,0])&(df1["set_1"]==df2.iloc[7,1])].sum()["par"]
j1=df1[(df1["group"]==df2.iloc[8,0])&(df1["set_1"]==df2.iloc[8,1])].sum()["val"]
j2=df1[(df1["group"]==df2.iloc[8,0])&(df1["set_1"]==df2.iloc[8,1])].sum()["par"]
k1=df1[(df1["group"]==df2.iloc[9,0])&(df1["set_2"]==df2.iloc[9,1])].sum()["val"]
k2=df1[(df1["group"]==df2.iloc[9,0])&(df1["set_2"]==df2.iloc[9,1])].sum()["par"]
l1=df1[(df1["group"]==df2.iloc[10,0])&(df1["set_2"]==df2.iloc[10,1])].sum()["val"]
l2=df1[(df1["group"]==df2.iloc[10,0])&(df1["set_2"]==df2.iloc[10,1])].sum()["par"]
m1=df1[(df1["group"]==df2.iloc[11,0])&(df1["set_4"]==df2.iloc[11,1])].sum()["val"]
m2=df1[(df1["group"]==df2.iloc[11,0])&(df1["set_4"]==df2.iloc[11,1])].sum()["par"]
n1=df1[(df1["group"]==df2.iloc[12,0])&(df1["set_4"]==df2.iloc[12,1])].sum()["val"]
n2=df1[(df1["group"]==df2.iloc[12,0])&(df1["set_4"]==df2.iloc[12,1])].sum()["par"]
Up to this part, I had to manually create "val" and "par" pairs for each element.
Then,
a=a1/a2
b=b1/b2
c=c1/c2
d=d1/d2
e=e1/e2
f=f1/f2
g=g1/g2
h=h1/h2
j=j1/j2
k=k1/k2
l=l1/l2
m=m1/m2
n=n1/n2
Finally, the result is:
df2["desired_calculation"]=[a,b,c,d,e,f,g,h,j,k,l,m,n]
python-3.x function apply
asked Nov 13 '18 at 11:48
nkltkfnkltkf
113
I have two data frames:
df1
df1 = pd.DataFrame( 'group' : ["A","A","A","B","B","B"],
'par' : [5,5,15,10,2,11],
'val' :[50,10,180,10,10,660],
'set_0' :["Country","Country","Country","Country","Country","Country"],
'set_1' :["size1","size1","size2","size3","size4","size3"],
'set_2' :["size12","size12","size12","size9","size13","size13"],
'set_3' :["size14","size14","size15","NO","NO","NO"],
'set_4' :["NO","NO","NO","size25","size25","size27"],
'set_5' :["NO","NO","NO","NO","NO","NO"]
)
df2
df2 = pd.DataFrame( 'group' : ["A","A","A","A","A","A","B","B","B","B","B","B","B"],
'set' : ["Country","size1","size2","size12","size14","size15","Country","size3","size4","size9","size13","size25","size27"],
)
For each row (group X set combination) of df2, I want to apply a function to calculate (sum of "val"/sum of "par").
I tried to use something with apply function but I could not really figure out since I am pretty new in python.
Could anyone please help regarding a solution?
below is the expected outcome:
As image upload keeps failing, I am also sharing below the code to get the outcome in the ugliest and most inefficient hardcoding way:
a1=df1[(df1["group"]==df2.iloc[0,0])&(df1["set_0"]==df2.iloc[0,1])].sum()["val"]
a2=df1[(df1["group"]==df2.iloc[0,0])&(df1["set_0"]==df2.iloc[0,1])].sum()["par"]
b1=df1[(df1["group"]==df2.iloc[1,0])&(df1["set_1"]==df2.iloc[1,1])].sum()["val"]
b2=df1[(df1["group"]==df2.iloc[1,0])&(df1["set_1"]==df2.iloc[1,1])].sum()["par"]
c1=df1[(df1["group"]==df2.iloc[2,0])&(df1["set_1"]==df2.iloc[2,1])].sum()["val"]
c2=df1[(df1["group"]==df2.iloc[2,0])&(df1["set_1"]==df2.iloc[2,1])].sum()["par"]
d1=df1[(df1["group"]==df2.iloc[3,0])&(df1["set_2"]==df2.iloc[3,1])].sum()["val"]
d2=df1[(df1["group"]==df2.iloc[3,0])&(df1["set_2"]==df2.iloc[3,1])].sum()["par"]
e1=df1[(df1["group"]==df2.iloc[4,0])&(df1["set_3"]==df2.iloc[4,1])].sum()["val"]
e2=df1[(df1["group"]==df2.iloc[4,0])&(df1["set_3"]==df2.iloc[4,1])].sum()["par"]
f1=df1[(df1["group"]==df2.iloc[5,0])&(df1["set_3"]==df2.iloc[5,1])].sum()["val"]
f2=df1[(df1["group"]==df2.iloc[5,0])&(df1["set_3"]==df2.iloc[5,1])].sum()["par"]
g1=df1[(df1["group"]==df2.iloc[6,0])&(df1["set_0"]==df2.iloc[6,1])].sum()["val"]
g2=df1[(df1["group"]==df2.iloc[6,0])&(df1["set_0"]==df2.iloc[6,1])].sum()["par"]
h1=df1[(df1["group"]==df2.iloc[7,0])&(df1["set_1"]==df2.iloc[7,1])].sum()["val"]
h2=df1[(df1["group"]==df2.iloc[7,0])&(df1["set_1"]==df2.iloc[7,1])].sum()["par"]
j1=df1[(df1["group"]==df2.iloc[8,0])&(df1["set_1"]==df2.iloc[8,1])].sum()["val"]
j2=df1[(df1["group"]==df2.iloc[8,0])&(df1["set_1"]==df2.iloc[8,1])].sum()["par"]
k1=df1[(df1["group"]==df2.iloc[9,0])&(df1["set_2"]==df2.iloc[9,1])].sum()["val"]
k2=df1[(df1["group"]==df2.iloc[9,0])&(df1["set_2"]==df2.iloc[9,1])].sum()["par"]
l1=df1[(df1["group"]==df2.iloc[10,0])&(df1["set_2"]==df2.iloc[10,1])].sum()["val"]
l2=df1[(df1["group"]==df2.iloc[10,0])&(df1["set_2"]==df2.iloc[10,1])].sum()["par"]
m1=df1[(df1["group"]==df2.iloc[11,0])&(df1["set_4"]==df2.iloc[11,1])].sum()["val"]
m2=df1[(df1["group"]==df2.iloc[11,0])&(df1["set_4"]==df2.iloc[11,1])].sum()["par"]
n1=df1[(df1["group"]==df2.iloc[12,0])&(df1["set_4"]==df2.iloc[12,1])].sum()["val"]
n2=df1[(df1["group"]==df2.iloc[12,0])&(df1["set_4"]==df2.iloc[12,1])].sum()["par"]
Up to this part, I had to manually create "val" and "par" pairs for each element.
Then,
a=a1/a2
b=b1/b2
c=c1/c2
d=d1/d2
e=e1/e2
f=f1/f2
g=g1/g2
h=h1/h2
j=j1/j2
k=k1/k2
l=l1/l2
m=m1/m2
n=n1/n2
Finally, the result is:
df2["desired_calculation"]=[a,b,c,d,e,f,g,h,j,k,l,m,n]
python-3.x function apply
python-3.x function apply
asked Nov 13 '18 at 11:48
nkltkfnkltkf
113
asked Nov 13 '18 at 11:48
nkltkfnkltkf
113
edited Nov 14 '18 at 10:31
nkltkf
asked Nov 13 '18 at 11:48
nkltkfnkltkf
113
asked Nov 13 '18 at 11:48
nkltkfnkltkf
113
asked Nov 13 '18 at 11:48
nkltkfnkltkf
113
113
Can you show us your desired output?
– jpp
Nov 13 '18 at 14:23
@jpp, not sure if the pasted output is gonna look readable, sharing it with an edit.
– nkltkf
Nov 13 '18 at 19:58
add a comment |
Can you show us your desired output?
– jpp
Nov 13 '18 at 14:23
@jpp, not sure if the pasted output is gonna look readable, sharing it with an edit.
– nkltkf
Nov 13 '18 at 19:58
Can you show us your desired output?
– jpp
Nov 13 '18 at 14:23
Can you show us your desired output?
– jpp
Nov 13 '18 at 14:23
@jpp, not sure if the pasted output is gonna look readable, sharing it with an edit.
– nkltkf
Nov 13 '18 at 19:58
@jpp, not sure if the pasted output is gonna look readable, sharing it with an edit.
– nkltkf
Nov 13 '18 at 19:58
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Can you show us your desired output?
– jpp
Nov 13 '18 at 14:23
@jpp, not sure if the pasted output is gonna look readable, sharing it with an edit.
– nkltkf
Nov 13 '18 at 19:58