I want to apply my custom function (it uses an if-else ladder) to these six columns (ERI_Hispanic, ERI_AmerInd_AKNatv, ERI_Asian, ERI_Black_Afr.Amer, ERI_HI_PacIsl, ERI_White) in each row of my dataframe.
I've tried different methods from other questions but still can't seem to find the right answer for my problem. The critical piece of this is that if the person is counted as Hispanic they can't be counted as anything else. Even if they have a "1" in another ethnicity column they still are counted as Hispanic not two or more races. Similarly, if the sum of all the ERI columns is greater than 1 they are counted as two or more races and can't be counted as a unique ethnicity(except for Hispanic). Hopefully this makes sense. Any help will be greatly appreciated.
Its almost like doing a for loop through each row and if each record meets a criterion they are added to one list and eliminated from the original.
From the dataframe below I need to calculate a new column based on the following spec in SQL:
========================= CRITERIA ===============================
IF [ERI_Hispanic] = 1 THEN RETURN “Hispanic”
ELSE IF SUM([ERI_AmerInd_AKNatv] + [ERI_Asian] + [ERI_Black_Afr.Amer] + [ERI_HI_PacIsl] + [ERI_White]) > 1 THEN RETURN “Two or More”
ELSE IF [ERI_AmerInd_AKNatv] = 1 THEN RETURN “A/I AK Native”
ELSE IF [ERI_Asian] = 1 THEN RETURN “Asian”
ELSE IF [ERI_Black_Afr.Amer] = 1 THEN RETURN “Black/AA”
ELSE IF [ERI_HI_PacIsl] = 1 THEN RETURN “Haw/Pac Isl.”
ELSE IF [ERI_White] = 1 THEN RETURN “White”
Comment: If the ERI Flag for Hispanic is True (1), the employee is classified as “Hispanic”
Comment: If more than 1 non-Hispanic ERI Flag is true, return “Two or More”
====================== DATAFRAME ===========================
lname fname rno_cd eri_afr_amer eri_asian eri_hawaiian eri_hispanic eri_nat_amer eri_white rno_defined
0 MOST JEFF E 0 0 0 0 0 1 White
1 CRUISE TOM E 0 0 0 1 0 0 White
2 DEPP JOHNNY 0 0 0 0 0 1 Unknown
3 DICAP LEO 0 0 0 0 0 1 Unknown
4 BRANDO MARLON E 0 0 0 0 0 0 White
5 HANKS TOM 0 0 0 0 0 1 Unknown
6 DENIRO ROBERT E 0 1 0 0 0 1 White
7 PACINO AL E 0 0 0 0 0 1 White
8 WILLIAMS ROBIN E 0 0 1 0 0 0 White
9 EASTWOOD CLINT E 0 0 0 0 0 1 White
The answers above are perfectly valid, but a vectorized solution exists, in the form of numpy.select. This allows you to define conditions, then define outputs for those conditions, much more efficiently than using apply:
First, define conditions:
conditions = [
df['eri_hispanic'] == 1,
df[['eri_afr_amer', 'eri_asian', 'eri_hawaiian', 'eri_nat_amer', 'eri_white']].sum(1).gt(1),
df['eri_nat_amer'] == 1,
df['eri_asian'] == 1,
df['eri_afr_amer'] == 1,
df['eri_hawaiian'] == 1,
df['eri_white'] == 1,
]
Now, define the corresponding outputs:
outputs = [
'Hispanic', 'Two Or More', 'A/I AK Native', 'Asian', 'Black/AA', 'Haw/Pac Isl.', 'White'
]
Finally, using numpy.select:
res = np.select(conditions, outputs, 'Other')
pd.Series(res)
0 White
1 Hispanic
2 White
3 White
4 Other
5 White
6 Two Or More
7 White
8 Haw/Pac Isl.
9 White
dtype: object
Why should numpy.select be used over apply? Here are some performance checks:
df = pd.concat([df]*1000)
In [42]: %timeit df.apply(lambda row: label_race(row), axis=1)
1.07 s ± 4.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [44]: %%timeit
...: conditions = [
...: df['eri_hispanic'] == 1,
...: df[['eri_afr_amer', 'eri_asian', 'eri_hawaiian', 'eri_nat_amer', 'eri_white']].sum(1).gt(1),
...: df['eri_nat_amer'] == 1,
...: df['eri_asian'] == 1,
...: df['eri_afr_amer'] == 1,
...: df['eri_hawaiian'] == 1,
...: df['eri_white'] == 1,
...: ]
...:
...: outputs = [
...: 'Hispanic', 'Two Or More', 'A/I AK Native', 'Asian', 'Black/AA', 'Haw/Pac Isl.', 'White'
...: ]
...:
...: np.select(conditions, outputs, 'Other')
...:
...:
3.09 ms ± 17 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Using numpy.select gives us vastly improved performance, and the discrepancy will only increase as the data grows.
try this,
df.loc[df['eri_white']==1,'race_label'] = 'White'
df.loc[df['eri_hawaiian']==1,'race_label'] = 'Haw/Pac Isl.'
df.loc[df['eri_afr_amer']==1,'race_label'] = 'Black/AA'
df.loc[df['eri_asian']==1,'race_label'] = 'Asian'
df.loc[df['eri_nat_amer']==1,'race_label'] = 'A/I AK Native'
df.loc[(df['eri_afr_amer'] + df['eri_asian'] + df['eri_hawaiian'] + df['eri_nat_amer'] + df['eri_white']) > 1,'race_label'] = 'Two Or More'
df.loc[df['eri_hispanic']==1,'race_label'] = 'Hispanic'
df['race_label'].fillna('Other', inplace=True)
O/P:
lname fname rno_cd eri_afr_amer eri_asian eri_hawaiian \
0 MOST JEFF E 0 0 0
1 CRUISE TOM E 0 0 0
2 DEPP JOHNNY NaN 0 0 0
3 DICAP LEO NaN 0 0 0
4 BRANDO MARLON E 0 0 0
5 HANKS TOM NaN 0 0 0
6 DENIRO ROBERT E 0 1 0
7 PACINO AL E 0 0 0
8 WILLIAMS ROBIN E 0 0 1
9 EASTWOOD CLINT E 0 0 0
eri_hispanic eri_nat_amer eri_white rno_defined race_label
0 0 0 1 White White
1 1 0 0 White Hispanic
2 0 0 1 Unknown White
3 0 0 1 Unknown White
4 0 0 0 White Other
5 0 0 1 Unknown White
6 0 0 1 White Two Or More
7 0 0 1 White White
8 0 0 0 White Haw/Pac Isl.
9 0 0 1 White White
use .loc instead of apply.
it improves vectorization.
.loc works in simple manner, mask rows based on the condition, apply values to the freeze rows.
for more details visit, .loc docs
Performance metrics:
Accepted Answer:
def label_race (row):
if row['eri_hispanic'] == 1 :
return 'Hispanic'
if row['eri_afr_amer'] + row['eri_asian'] + row['eri_hawaiian'] + row['eri_nat_amer'] + row['eri_white'] > 1 :
return 'Two Or More'
if row['eri_nat_amer'] == 1 :
return 'A/I AK Native'
if row['eri_asian'] == 1:
return 'Asian'
if row['eri_afr_amer'] == 1:
return 'Black/AA'
if row['eri_hawaiian'] == 1:
return 'Haw/Pac Isl.'
if row['eri_white'] == 1:
return 'White'
return 'Other'
df=pd.read_csv('dataser.csv')
df = pd.concat([df]*1000)
%timeit df.apply(lambda row: label_race(row), axis=1)
1.15 s ± 46.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
My Proposed Answer:
def label_race(df):
df.loc[df['eri_white']==1,'race_label'] = 'White'
df.loc[df['eri_hawaiian']==1,'race_label'] = 'Haw/Pac Isl.'
df.loc[df['eri_afr_amer']==1,'race_label'] = 'Black/AA'
df.loc[df['eri_asian']==1,'race_label'] = 'Asian'
df.loc[df['eri_nat_amer']==1,'race_label'] = 'A/I AK Native'
df.loc[(df['eri_afr_amer'] + df['eri_asian'] + df['eri_hawaiian'] + df['eri_nat_amer'] + df['eri_white']) > 1,'race_label'] = 'Two Or More'
df.loc[df['eri_hispanic']==1,'race_label'] = 'Hispanic'
df['race_label'].fillna('Other', inplace=True)
df=pd.read_csv('s22.csv')
df = pd.concat([df]*1000)
%timeit label_race(df)
24.7 ms ± 1.7 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
.apply() takes in a function as the first parameter; pass in the label_race function as so:
df['race_label'] = df.apply(label_race, axis=1)
You don't need to make a lambda function to pass in a function.
As @user3483203 pointed out, numpy.select is the best approach
Store your conditional statements and the corresponding actions in two lists
conds = [(df['eri_hispanic'] == 1),(df[['eri_afr_amer', 'eri_asian', 'eri_hawaiian', 'eri_nat_amer', 'eri_white']].sum(1).gt(1)),(df['eri_nat_amer'] == 1),(df['eri_asian'] == 1),(df['eri_afr_amer'] == 1),(df['eri_hawaiian'] == 1),(df['eri_white'] == 1,])
actions = ['Hispanic', 'Two Or More', 'A/I AK Native', 'Asian', 'Black/AA', 'Haw/Pac Isl.', 'White']
You can now use np.select using these lists as its arguments
df['label_race'] = np.select(conds,actions,default='Other')
Reference: https://numpy.org/doc/stable/reference/generated/numpy.select.html
Since this is the first Google result for 'pandas new column from others', here's a simple example:
import pandas as pd
# make a simple dataframe
df = pd.DataFrame({'a':[1,2], 'b':[3,4]})
df
# a b
# 0 1 3
# 1 2 4
# create an unattached column with an index
df.apply(lambda row: row.a + row.b, axis=1)
# 0 4
# 1 6
# do same but attach it to the dataframe
df['c'] = df.apply(lambda row: row.a + row.b, axis=1)
df
# a b c
# 0 1 3 4
# 1 2 4 6
If you get the SettingWithCopyWarning you can do it this way also:
fn = lambda row: row.a + row.b # define a function for the new column
col = df.apply(fn, axis=1) # get column data with an index
df = df.assign(c=col.values) # assign values to column 'c'
Source: https://stackoverflow.com/a/12555510/243392
And if your column name includes spaces you can use syntax like this:
df = df.assign(**{'some column name': col.values})
Source: Stackoverflow.com