I'm trying to create a reusable function in python 2.7(pandas) to form categorical bins, i.e. group less-value categories as 'other'. Can someone help me to create a function for the below: col1, col2, etc. are different categorical variable columns.
##Reducing categories by binning categorical variables - column1
a = df.col1.value_counts()
#get top 5 values of index
vals = a[:5].index
df['col1_new'] = df.col1.where(df.col1.isin(vals), 'other')
df = df.drop(['col1'],axis=1)
##Reducing categories by binning categorical variables - column2
a = df.col2.value_counts()
#get top 6 values of index
vals = a[:6].index
df['col2_new'] = df.col2.where(df.col2.isin(vals), 'other')
df = df.drop(['col2'],axis=1)
You can use:
df = pd.DataFrame({'A':list('abcdefabcdefabffeg'),
'D':[1,3,5,7,1,0,1,3,5,7,1,0,1,3,5,7,1,0]})
print (df)
A D
0 a 1
1 b 3
2 c 5
3 d 7
4 e 1
5 f 0
6 a 1
7 b 3
8 c 5
9 d 7
10 e 1
11 f 0
12 a 1
13 b 3
14 f 5
15 f 7
16 e 1
17 g 0
def replace_under_top(df, c, n):
a = df[c].value_counts()
#get top n values of index
vals = a[:n].index
#assign columns back
df[c] = df[c].where(df[c].isin(vals), 'other')
#rename processes column
df = df.rename(columns={c : c + '_new'})
return df
Test:
df1 = replace_under_top(df, 'A', 3)
print (df1)
A_new D
0 other 1
1 b 3
2 other 5
3 other 7
4 e 1
5 f 0
6 other 1
7 b 3
8 other 5
9 other 7
10 e 1
11 f 0
12 other 1
13 b 3
14 f 5
15 f 7
16 e 1
17 other 0
df2 = replace_under_top(df, 'D', 4)
print (df2)
A D_new
0 other 1
1 b 3
2 other 5
3 other 7
4 e 1
5 f other
6 other 1
7 b 3
8 other 5
9 other 7
10 e 1
11 f other
12 other 1
13 b 3
14 f 5
15 f 7
16 e 1
17 other other
Related
I want to use dataframe.melt function in pandas lib to convert data format from rows into column but keeping first column value. I ve just tried also .pivot, but it is not working good. Please look at the example below and please help:
ID Alphabet Unspecified: 1 Unspecified: 2
0 1 A G L
1 2 B NaN NaN
2 3 C H NaN
3 4 D I M
4 5 E J NaN
5 6 F K O
Into this:
ID Alphabet
0 1 A
1 1 G
2 1 L
3 2 B
4 3 C
5 3 H
6 4 D
7 4 I
8 4 M
9 5 E
10 5 J
11 6 F
12 6 K
11 6 O
Try (assuming ID is unique and sorted):
df = (
pd.melt(df, "ID")
.sort_values("ID", kind="stable")
.drop(columns="variable")
.dropna()
.reset_index(drop=True)
.rename(columns={"value": "Alphabet"})
)
print(df)
Prints:
ID Alphabet
0 1 A
1 1 G
2 1 L
3 2 B
4 3 C
5 3 H
6 4 D
7 4 I
8 4 M
9 5 E
10 5 J
11 6 F
12 6 K
13 6 O
Don't melt but rather stack, this will directly drop the NaNs and keep the order per row:
out = (df
.set_index('ID')
.stack().droplevel(1)
.reset_index(name='Alphabet')
)
Output:
ID Alphabet
0 1 A
1 1 G
2 1 L
3 2 B
4 3 C
5 3 H
6 4 D
7 4 I
8 4 M
9 5 E
10 5 J
11 6 F
12 6 K
13 6 O
One option is with pivot_longer from pyjanitor:
# pip install pyjanitor
import pandas as pd
import janitor
(df
.pivot_longer(
index = 'ID',
names_to = 'Alphabet',
names_pattern = ['.+'],
sort_by_appearance = True)
.dropna()
)
ID Alphabet
0 1 A
1 1 G
2 1 L
3 2 B
6 3 C
7 3 H
9 4 D
10 4 I
11 4 M
12 5 E
13 5 J
15 6 F
16 6 K
17 6 O
In the code above, the names_pattern accepts a list of regular expression to match the desired columns, all the matches are collated into one column names Alphabet in names_to.
Suppose I have the following Pandas dataframe:
In[285]: df = pd.DataFrame({'Name':['A','B'], 'Start': [1,6], 'End': [4,12]})
In [286]: df
Out[286]:
Name Start End
0 A 1 4
1 B 6 12
Now I would like to construct the dataframe as follows:
Name Number
0 A 1
1 A 2
2 A 3
3 A 4
4 B 6
5 B 7
6 B 8
7 B 9
8 B 10
9 B 11
10 B 12
My biggest struggle is in getting the 'Name' column right. Is there a smart way to do this in Python?
I would do pd.concat on a list comprehension:
pd.concat(pd.DataFrame({'Number': np.arange(s,e+1)})
.assign(Name=n)
for n,s,e in zip(df['Name'], df['Start'], df['End']))
Output:
Number Name
0 1 A
1 2 A
2 3 A
3 4 A
0 6 B
1 7 B
2 8 B
3 9 B
4 10 B
5 11 B
6 12 B
Update: As commented by #rafaelc:
pd.concat(pd.DataFrame({'Number': np.arange(s,e+1), 'Name': n})
for n,s,e in zip(df['Name'], df['Start'], df['End']))
works just fine.
Let us do it with this example (with 3 names):
import pandas as pd
df = pd.DataFrame({'Name':['A','B','C'], 'Start': [1,6,18], 'End': [4,12,20]})
You may create the target columns first, using list comprehensions:
name = [row.Name for i, row in df.iterrows() for _ in range(row.End - row.Start + 1)]
number = [k for i, row in df.iterrows() for k in range(row.Start, row.End + 1)]
And then you can create the target DataFrame:
expanded = pd.DataFrame({"Name": name, "Number": number})
You get:
Name Number
0 A 1
1 A 2
2 A 3
3 A 4
4 B 6
5 B 7
6 B 8
7 B 9
8 B 10
9 B 11
10 B 12
11 C 18
12 C 19
13 C 20
I'd take advantage of loc and index.repeat for a vectorized solution.
base = df.loc[df.index.repeat(df['End'] - df['Start'] + 1), ['Name', 'Start']]
base['Start'] += base.groupby(level=0).cumcount()
Name Start
0 A 1
0 A 2
0 A 3
0 A 4
1 B 6
1 B 7
1 B 8
1 B 9
1 B 10
1 B 11
1 B 12
Of course we can rename the columns and reset the index at the end, for a nicer showing.
base.rename(columns={'Start': 'Number'}).reset_index(drop=True)
Name Number
0 A 1
1 A 2
2 A 3
3 A 4
4 B 6
5 B 7
6 B 8
7 B 9
8 B 10
9 B 11
10 B 12
Say I have a row of column headers, and associated values in a Pandas Dataframe:
print df
A B C D E F G H I J K
1 2 3 4 5 6 7 8 9 10 11
how do I go about displaying them like the following:
print df
A B C D E
1 2 3 4 5
F G H I J
6 7 8 9 10
K
11
custom function
def new_repr(self):
g = self.groupby(np.arange(self.shape[1]) // 5, axis=1)
return '\n\n'.join([d.to_string() for _, d in g])
print(new_repr(df))
A B C D E
0 1 2 3 4 5
F G H I J
0 6 7 8 9 10
K
0 11
pd.set_option('display.width', 20)
pd.set_option('display.expand_frame_repr', True)
df
A B C D E \
0 1 2 3 4 5
F G H I J \
0 6 7 8 9 10
K
0 11
Consider the following dataframe:
df = pd.DataFrame(data=np.array([['a',1, 2, 3,'T'], ['a',4, 5, 6,'F'],
['b',7, 8, 9,'T'], ['b',10, 11 , 12,'T'], ['b',13, 14 , 15,'F']])
, columns=['id','A', 'B', 'C','T/F'])
id A B C T/F
0 a 1 2 3 T
1 a 4 5 6 F
2 b 7 8 9 T
3 b 10 11 12 T
4 b 13 14 15 F
I want to apply a condition to T/F column that will copy the rows of each id with T label to further columns of the same id.
For example,I need the following result:
id A B C T/F
0 a 1 2 3 T 1 2 3 T
1 a 4 5 6 F 1 2 3 T
2 b 7 8 9 T 7 8 9 T
3 b 10 11 12 T 7 8 9 T
4 b 13 14 15 F 7 8 9 T
5 b 7 8 9 T 10 11 12 T
6 b 10 11 12 T 10 11 12 T
7 b 13 14 15 F 10 11 12 T
here is my script:
n = np.array(df.groupby('id').size())
m = len(df.groupby('id'))
Cnt = 0
df4 = pd.DataFrame()
for prsnNo in range(m):
for i in range(n[prsnNo]):
v = df.iloc[Cnt: Cnt + n[prsnNo], :].groupby('id').cumcount() == i
df1 = df.iloc[Cnt: Cnt + n[prsnNo], :].where(v)
temp = df4
df4 = df.iloc[Cnt: Cnt + n[prsnNo], :].merge(df1, on="id", how="left")
df4 = pd.concat([temp, df4])
Cnt += n[prsnNo]
I do not know how to add a condition to check the value of T/F column in my loop.If I add the if condition in my loop it gives me an error.
for prsnNo in range(m):
for i in range(n[prsnNo]):
if df[df['T/F'] =='T'] :
v = df.iloc[Cnt: Cnt + n[prsnNo], :].groupby('id').cumcount() == i
df1 = df.iloc[Cnt: Cnt + n[prsnNo], :].where(v)
temp = df4
df4 = df.iloc[Cnt: Cnt + n[prsnNo], :].merge(df1, on="id", how="left")
df4 = pd.concat([temp, df4])
Cnt += n[prsnNo]
Thanks,
If order doesn't matter, you can use groupby + first, and then perform a merge with df and the grouped result.
v = df.groupby(['id', df['T/F'].eq('T').cumsum()])\
.first().reset_index(level=1, drop=True)
df = df.merge(v, left_on='id', right_index=True)
df.columns = df.columns.str.split('_').str[0]
df
id A B C T/F A B C T/F
0 a 1 2 3 T 1 2 3 T
1 a 4 5 6 F 1 2 3 T
2 b 7 8 9 T 7 8 9 T
2 b 7 8 9 T 10 11 12 T
3 b 10 11 12 T 7 8 9 T
3 b 10 11 12 T 10 11 12 T
4 b 13 14 15 F 7 8 9 T
4 b 13 14 15 F 10 11 12 T
I have an existing pandas DataFrame, and I want to add a new column, where the value of each row will depend on the previous row.
for example:
df1 = pd.DataFrame(np.random.randint(10, size=(4, 4)), columns=['a', 'b', 'c', 'd'])
df1
Out[31]:
a b c d
0 9 3 3 0
1 3 9 5 1
2 1 7 5 6
3 8 0 1 7
and now I want to create column e, where for each row i the value of df1['e'][i] would be: df1['e'][i] = df1['d'][i] - df1['d'][i-1]
desired output:
df1:
a b c d e
0 9 3 3 0 0
1 3 9 5 1 1
2 1 7 5 6 5
3 8 0 1 7 1
how can I achieve this?
You can use sub with shift:
df['e'] = df.d.sub(df.d.shift(), fill_value=0)
print (df)
a b c d e
0 9 3 3 0 0.0
1 3 9 5 1 1.0
2 1 7 5 6 5.0
3 8 0 1 7 1.0
If need convert to int:
df['e'] = df.d.sub(df.d.shift(), fill_value=0).astype(int)
print (df)
a b c d e
0 9 3 3 0 0
1 3 9 5 1 1
2 1 7 5 6 5
3 8 0 1 7 1