1

df

        Date    Factory I1  D1  I2  D2  I3  D3

   0    1701    West    0   0   0   0   10  0

   1    1701    East    39  0   83  15  3   15

   2    1701    West    0   0   0   10  15  0

I want this:

 If D2>I2 Then I3=I2+I3,D3=D2+D3,I2=D2=0

 If D3>I3 Then I2=I2+I3,D2=D2+D3,I3=D3=0

else no change

So the data should be like this:

df

     Date   Factory I1  D1  I2  D2  I3  D3

     0  1701    West    0   0   0   0   10  0

     1  1701    East    39  0   86  30  0   0

     2  1701    West    0   0   0   0   15  10

How can I do?Please help~

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2 Answers 2

Reset to default
2

You need mask with assign:

m1 = df.D2 > df.I2
m2 = df.D3 > df.I3
df = df.mask(m1, df.assign(I3=df.I2+df.I3, D3=df.D2+df.D3, I2=0, D2=0))
df = df.mask(m2, df.assign(I2=df.I2+df.I3, D2=df.D2+df.D3, I3=0, D3=0))
print (df)
   Date Factory  I1  D1  I2  D2  I3  D3
0  1701    West   0   0   0   0  10   0
1  1701    East  39   0  86  30   0   0
2  1701    West   0   0   0   0  15  10

If there are columns with spaces:

m1 = df.D2 > df.I2
m2 = df.D3 > df.I3

def f1(x):
    df['I3']=df.I2+df.I3 
    df['D3']=df.D2+df.D3 
    df['I2'] = df['D2'] = 0
    return df

def f2(x):
    df['I2']=df.I2+df.I3 
    df['D2']=df.D2+df.D3 
    df['I3'] = df['D3'] = 0
    return df

df = df.mask(m1, f1)
df = df.mask(m2, f2)
print (df)
   Date Factory  I1  D1  I2  D2  I3  D3
0  1701    West   0   0  10   0   0   0
1  1701    East  39   0  86  30   0   0
2  1701    West   0   0  15  10   0   0

Or if want use assign (I3 was changed to I 3):

m1 = df['D2'] > df['I2']
m2 = df['D3'] > df['I 3']

d1 = {'I 3': lambda x: x['I2'] + x['I 3'],'D3':lambda x: x['D2']+x['D3'],'I2':0,'D2':0}
d2 = {'I2': lambda x: x['I2'] + x['I 3'],'D2':lambda x: x['D2']+x['D3'],'I3':0,'D3':0}
df = df.mask(m1, df.assign(**d1))
df = df.mask(m2, df.assign(**d2))
print (df)
   Date Factory  I1  D1  I2  D2  I 3  D3
0  1701    West   0   0   0   0   10   0
1  1701    East  39   0  86  30    3   0
2  1701    West   0   0   0   0   15  10
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1 Comment

another question：Now the columns names are like I3，D3. When they come to be Input third ， Defect third，how can this work？ there is a space in columns name…
0

One can use easily understandable for loop and work over each row: 

newvals = []
for i in range(len(df.index)):
    ROW = df.iloc[i,:]
    if ROW.D2>ROW.I2:
        ROW.I3=ROW.I2+ROW.I3
        ROW.D3=ROW.D2+ROW.D3
        ROW.I2=ROW.D2=0 
    if ROW.D3>ROW.I3:
        ROW.I2=ROW.I2+ROW.I3
        ROW.D2=ROW.D2+ROW.D3
        ROW.I3=ROW.D3=0
    newvals.append(list(ROW))
newdf = pd.DataFrame(data=newvals, columns=df.columns)
print(newdf)

Output:

   Date Factory  I1  D1  I2  D2  I3  D3
0  1701    West   0   0   0   0  10   0
1  1701    East  39   0  86  30   0   0
2  1701    West   0   0   0   0  15  10
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3 Comments

In my opinion possible solution, but the best in pandas is avoid loops, because slow.
Yes，it’s easier to understand but as jezrael says，it’s not fast. TKU all the same！
What did you mean by 'one by one' in your title?

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