Python pandas. Replace values in columns

Question

Suppose i have a DataFrame objects like this:

age_diff    result
       1         0
      -1         1
       0         1

I want to replace negative values in column age_diff by applying to them abs function. Also, if value in age_diff is changed, value in column result should be switched(if it was 0 then to 1, else to 0).

After this transformation DataFrame, shown above, should look like:

age_diff    result
       1         0
       1         0
       0         1

Can, anyone, point me to how can i achieve this?

jezrael · Accepted Answer · 2017-05-25 08:08:01Z

5

Use abs + mask + map by dict:

print (df)
   age_diff  result
0         1       0
1        -1       0
2        -1       1
3         0       1


mask = df['age_diff'] < 0
df['age_diff'] = df['age_diff'].abs()
df['result'] = df['result'].mask(mask, df['result'].map({0:1, 1:0}))
print (df)
   age_diff  result
0         1       0
1         1       1
2         1       0
3         0       1

Another solution - instead map cast to bool and invert by ~:

mask = df['age_diff'] < 0
df['age_diff'] = df['age_diff'].abs()
df['result'] = df['result'].mask(mask, ~(df['result'].astype(bool)))
print (df)
   age_diff  result
0         1       0
1         1       1
2         1       0
3         0       1

edited May 25, 2017 at 8:08

answered May 25, 2017 at 7:56

jezrael

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1 Comment

jezrael Over a year ago

Super, glad can help. Nice day!

AndreyF · Accepted Answer · 2017-05-25 08:55:43Z

2

Another approach you can take is applying a function:

def magic(row):
    if abs(row['age_diff']) != row['age_diff']:
        row['age_diff'] = abs(row['age_diff'])
        row['result'] = 1 - row['result']
    return row

df = df.apply(magic,axis=1)

answered May 25, 2017 at 8:55

AndreyF

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Comments

piRSquared · Accepted Answer · 2017-05-25 09:30:39Z

1

I get the signs of age_diff with np.sign and manipulate from there.

s = np.sign(df.age_diff.values)

df.age_diff *= s
df.result -= (s == -1)

print(df)

   age_diff  result
0         1       0
1         1       0
2         0       1

answered May 25, 2017 at 9:30

piRSquared

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Comments

Allen Qin · Accepted Answer · 2017-05-25 12:36:20Z

1

Another one liner solution using lambda to get abs of age_diff and then change sign of result if age_diff is negative.

df.apply(lambda x: [abs(x.age_diff),abs((x.age_diff<0) - x.result)],axis=1)
Out[165]: 
   age_diff  result
0         1       0
1         1       0
2         0       1

answered May 25, 2017 at 12:36

Allen Qin

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