Pandas Dataframe: create new label columns based on other columns
I have this example pandas.DataFrame
with + 20K lines, in the following form:
import pandas as pd
import numpy as np
data = {"first_column": ["A", "B", "B", "B", "C", "A", "A", "A", "D", "B", "A", "A"],
"second_column": [0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0]}
df = pd.DataFrame(data)
>>> df
first_column second_column
0 A 0
1 B 1
2 B 1
3 B 1
4 C 0
5 A 0
6 A 0
7 A 1
8 D 1
9 B 1
10 A 1
11 A 0
....
The column first_column
has for each row A
, B
, C
and D
. The second column contains a binary label that represents a group of values. All consecutive groups of 1 are a unique "group", for example. lines 1-3 are one group, lines 7-10 are another group.
I would like to "label" each of these groups by being either "AB" (the group consists of only A or B), "CD" (the group consists of only C or D), or "mixed" (if there is a mixture, for example, all B and one C). It would also be useful to know "how" some of these groupings are mixed, ie. Percentage AB of total labels. So, if it's only A
or B
, then the identifier should be AB
. If it is only C
or D
, then the identifier must be CD
. It's a mixture of A, B, C and / or D, then she is mixed
. Percentage (# of AB lines) / (# of all lines)
This is what the result would look like DataFrame
:
>>> df
first_column second_column identity percent
0 A 0 0 0
1 B 1 AB 1.0
2 B 1 AB 1.0
3 B 1 AB 1.0
4 C 0 0 0
5 A 0 0 0
6 A 0 0 0
7 A 1 mixed 0.75 # 3/4, 3-AB, 4-total
8 D 1 mixed 0.75
9 B 1 mixed 0.75
10 A 1 mixed 0.75
11 A 0 0 0
....
My initial thought was to first try to use df.loc()
with
if (df.first_column == "A" | df.first_column == "B"):
df.loc[df.second_column == 1, "identity"] = "AB"
if (df.first_column == "C" | df.first_column == "D"):
df.loc[df.second_column == 1, "identity"] = "CD"
but this does not account for mixtures and does not work with isolated groups.
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Here's one way to do it.
Code:
import pandas as pd
from collections import Counter
a_b = set('AB')
c_d = set('CD')
def get_id_percent(group):
present = Counter(group['first_column'])
present_set = set(present.keys())
if group['second_column'].iloc[0] == 0:
ret_val = 0, 0
elif present_set.issubset(a_b) and len(present_set) == 1:
ret_val = 'AB', 0
elif present_set.issubset(c_d) and len(present_set) == 1:
ret_val = 'CD', 0
else:
ret_val = 'mixed', \
float(present['A'] + present['B']) / len(group)
return pd.DataFrame(
[ret_val] * len(group), columns=['identity', 'percent'])
Test code:
data = {"first_column": ["A", "B", "B", "B", "C", "A", "A",
"A", "D", "B", "A", "A"],
"second_column": [0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0]}
df = pd.DataFrame(data)
groupby = df.groupby((df.second_column != df.second_column.shift()).cumsum())
results = groupby.apply(get_id_percent).reset_index()
results = results.drop(['second_column', 'level_1'], axis=1)
df = pd.concat([df, results], axis=1)
print(df)
Results:
first_column second_column identity percent
0 A 0 0 0.00
1 B 1 AB 0.00
2 B 1 AB 0.00
3 B 1 AB 0.00
4 C 0 0 0.00
5 A 0 0 0.00
6 A 0 0 0.00
7 A 1 mixed 0.75
8 D 1 mixed 0.75
9 B 1 mixed 0.75
10 A 1 mixed 0.75
11 A 0 0 0.00
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Here's one approach:
import pandas as pd
# generate example data
data = {"first_column": ["A", "B", "B", "B", "C", "A", "A", "A", "D", "B", "A", "A"],
"second_column": [0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0]}
df = pd.DataFrame(data)
# these are intermediary groups for computation
df['group_type'] = None
df['ct'] = 0
def find_border(x, ct):
''' finds and labels lettered groups '''
ix = x.name
# does second_column == 1?
if x.second_column:
# if it the start of a group...
if (not ix) | (not df.group_type[ix-1]):
df.ix[ix,'group_type'] = x.first_column
df.ix[ix,'ct'] += 1
return
# if it the end of a group
elif (not df.second_column[ix+1]):
df.ix[ix,'group_type'] = df.group_type[ix-1] + x.first_column
df.ix[ix,'ct'] = df.ct[ix-1] + 1
for i in range(df.ct[ix-1]+1):
df.ix[ix-i,'group_type'] = df.ix[ix,'group_type']
df.ix[ix,'ct'] = 0
return
# if it the middle of a group
else:
df.ix[ix,'ct'] = df.ct[ix-1] + 1
df.ix[ix,'group_type'] = df.group_type[ix-1] + x.first_column
return
return
# compute group membership
_=df.apply(find_border, axis='columns', args=(0,))
def determine_id(x):
if not x:
return '0'
if list(set(x)) in [['A'],['B'],['A','B']]:
return 'AB'
elif list(set(x)) in [['C'],['D'],['C','D']]:
return 'CD'
else:
return 'mixed'
def determine_pct(x):
if not x:
return 0
return sum([1 for letter in x if letter in ['A','B']]) / float(len(x))
# determine row identity
df['identity'] = df.group_type.apply(determine_id)
# determine % of A or B in group
df['percent'] = df.group_type.apply(determine_pct)
Output:
first_column second_column identity percent
0 A 0 0 0.00
1 B 1 AB 1.00
2 B 1 AB 1.00
3 B 1 AB 1.00
4 C 0 0 0.00
5 A 0 0 0.00
6 A 0 0 0.00
7 A 1 mixed 0.75
8 D 1 mixed 0.75
9 B 1 mixed 0.75
10 A 1 mixed 0.75
11 A 0 0 0.00
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