A pythonic and uFunc-y way to turn pandas column into “increasing” index?
up vote
6
down vote
favorite
Let's say I have a pandas df like so:
Index A B
0 foo 3
1 foo 2
2 foo 5
3 bar 3
4 bar 4
5 baz 5
What's a good fast way to add a column like so:
Index A B Aidx
0 foo 3 0
1 foo 2 0
2 foo 5 0
3 bar 3 1
4 bar 4 1
5 baz 5 2
I.e. adding an increasing index for each unique value?
I know I could use df.unique()
, then use a dict and enumerate
to create a lookup, and then apply that dictionary lookup to create the column. But I feel like there should be faster way, possibly involving groupby
with some special function?
python pandas
add a comment |
up vote
6
down vote
favorite
Let's say I have a pandas df like so:
Index A B
0 foo 3
1 foo 2
2 foo 5
3 bar 3
4 bar 4
5 baz 5
What's a good fast way to add a column like so:
Index A B Aidx
0 foo 3 0
1 foo 2 0
2 foo 5 0
3 bar 3 1
4 bar 4 1
5 baz 5 2
I.e. adding an increasing index for each unique value?
I know I could use df.unique()
, then use a dict and enumerate
to create a lookup, and then apply that dictionary lookup to create the column. But I feel like there should be faster way, possibly involving groupby
with some special function?
python pandas
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let's say I have a pandas df like so:
Index A B
0 foo 3
1 foo 2
2 foo 5
3 bar 3
4 bar 4
5 baz 5
What's a good fast way to add a column like so:
Index A B Aidx
0 foo 3 0
1 foo 2 0
2 foo 5 0
3 bar 3 1
4 bar 4 1
5 baz 5 2
I.e. adding an increasing index for each unique value?
I know I could use df.unique()
, then use a dict and enumerate
to create a lookup, and then apply that dictionary lookup to create the column. But I feel like there should be faster way, possibly involving groupby
with some special function?
python pandas
Let's say I have a pandas df like so:
Index A B
0 foo 3
1 foo 2
2 foo 5
3 bar 3
4 bar 4
5 baz 5
What's a good fast way to add a column like so:
Index A B Aidx
0 foo 3 0
1 foo 2 0
2 foo 5 0
3 bar 3 1
4 bar 4 1
5 baz 5 2
I.e. adding an increasing index for each unique value?
I know I could use df.unique()
, then use a dict and enumerate
to create a lookup, and then apply that dictionary lookup to create the column. But I feel like there should be faster way, possibly involving groupby
with some special function?
python pandas
python pandas
asked 1 hour ago
Lagerbaer
2,6181124
2,6181124
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
One way is to use ngroup
. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False
:
df['Aidx'] = df.groupby('A',sort=False).ngroup()
>>> df
Index A B Aidx
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
add a comment |
up vote
2
down vote
No need groupby
using
Method 1factorize
pd.factorize(df.A)[0]
array([0, 0, 0, 1, 1, 2], dtype=int64)
#df['Aidx']=pd.factorize(df.A)[0]
Method 2 sklearn
from sklearn import preprocessing
le = preprocessing.LabelEncoder()
le.fit(df.A)
LabelEncoder()
le.transform(df.A)
array([2, 2, 2, 0, 0, 1])
Method 3 cat.codes
df.A.astype('category').cat.codes
Method 4 map
+ unique
l=df.A.unique()
df.A.map(dict(zip(l,range(len(l)))))
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int64
add a comment |
up vote
2
down vote
One more method of doing so could be.
df['C'] = i.ne(df.A.shift()).cumsum()-1
df
When we print df
value it will be as follows.
Index A B C
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
Explanation of solution: Let's break above solution into parts for understanding purposes.
1st step: Compare df's A column by shifting its value down to itself as follows.
i.ne(df.A.shift())
Output we will get is:
0 True
1 False
2 False
3 True
4 False
5 True
2nd step: Use of cumsum()
function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum()
function and its value will be increased.
i.ne(df.A.shift()).cumsum()-1
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int32
3rd step: Save command's value into df['C']
which will create a new column named C
in df
.
1
Nice method ve++ for you
– W-B
36 mins ago
@W-B, thank you for encouragement sir, ++ve for your unique style already :)
– RavinderSingh13
34 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
One way is to use ngroup
. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False
:
df['Aidx'] = df.groupby('A',sort=False).ngroup()
>>> df
Index A B Aidx
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
add a comment |
up vote
6
down vote
One way is to use ngroup
. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False
:
df['Aidx'] = df.groupby('A',sort=False).ngroup()
>>> df
Index A B Aidx
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
add a comment |
up vote
6
down vote
up vote
6
down vote
One way is to use ngroup
. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False
:
df['Aidx'] = df.groupby('A',sort=False).ngroup()
>>> df
Index A B Aidx
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
One way is to use ngroup
. Just remember you have to make sure your groupby isn't resorting the groups to get your desired output, so set sort=False
:
df['Aidx'] = df.groupby('A',sort=False).ngroup()
>>> df
Index A B Aidx
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
edited 19 mins ago
answered 1 hour ago
sacul
29.7k41640
29.7k41640
add a comment |
add a comment |
up vote
2
down vote
No need groupby
using
Method 1factorize
pd.factorize(df.A)[0]
array([0, 0, 0, 1, 1, 2], dtype=int64)
#df['Aidx']=pd.factorize(df.A)[0]
Method 2 sklearn
from sklearn import preprocessing
le = preprocessing.LabelEncoder()
le.fit(df.A)
LabelEncoder()
le.transform(df.A)
array([2, 2, 2, 0, 0, 1])
Method 3 cat.codes
df.A.astype('category').cat.codes
Method 4 map
+ unique
l=df.A.unique()
df.A.map(dict(zip(l,range(len(l)))))
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int64
add a comment |
up vote
2
down vote
No need groupby
using
Method 1factorize
pd.factorize(df.A)[0]
array([0, 0, 0, 1, 1, 2], dtype=int64)
#df['Aidx']=pd.factorize(df.A)[0]
Method 2 sklearn
from sklearn import preprocessing
le = preprocessing.LabelEncoder()
le.fit(df.A)
LabelEncoder()
le.transform(df.A)
array([2, 2, 2, 0, 0, 1])
Method 3 cat.codes
df.A.astype('category').cat.codes
Method 4 map
+ unique
l=df.A.unique()
df.A.map(dict(zip(l,range(len(l)))))
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int64
add a comment |
up vote
2
down vote
up vote
2
down vote
No need groupby
using
Method 1factorize
pd.factorize(df.A)[0]
array([0, 0, 0, 1, 1, 2], dtype=int64)
#df['Aidx']=pd.factorize(df.A)[0]
Method 2 sklearn
from sklearn import preprocessing
le = preprocessing.LabelEncoder()
le.fit(df.A)
LabelEncoder()
le.transform(df.A)
array([2, 2, 2, 0, 0, 1])
Method 3 cat.codes
df.A.astype('category').cat.codes
Method 4 map
+ unique
l=df.A.unique()
df.A.map(dict(zip(l,range(len(l)))))
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int64
No need groupby
using
Method 1factorize
pd.factorize(df.A)[0]
array([0, 0, 0, 1, 1, 2], dtype=int64)
#df['Aidx']=pd.factorize(df.A)[0]
Method 2 sklearn
from sklearn import preprocessing
le = preprocessing.LabelEncoder()
le.fit(df.A)
LabelEncoder()
le.transform(df.A)
array([2, 2, 2, 0, 0, 1])
Method 3 cat.codes
df.A.astype('category').cat.codes
Method 4 map
+ unique
l=df.A.unique()
df.A.map(dict(zip(l,range(len(l)))))
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int64
edited 28 mins ago
answered 38 mins ago
W-B
97.8k73162
97.8k73162
add a comment |
add a comment |
up vote
2
down vote
One more method of doing so could be.
df['C'] = i.ne(df.A.shift()).cumsum()-1
df
When we print df
value it will be as follows.
Index A B C
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
Explanation of solution: Let's break above solution into parts for understanding purposes.
1st step: Compare df's A column by shifting its value down to itself as follows.
i.ne(df.A.shift())
Output we will get is:
0 True
1 False
2 False
3 True
4 False
5 True
2nd step: Use of cumsum()
function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum()
function and its value will be increased.
i.ne(df.A.shift()).cumsum()-1
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int32
3rd step: Save command's value into df['C']
which will create a new column named C
in df
.
1
Nice method ve++ for you
– W-B
36 mins ago
@W-B, thank you for encouragement sir, ++ve for your unique style already :)
– RavinderSingh13
34 mins ago
add a comment |
up vote
2
down vote
One more method of doing so could be.
df['C'] = i.ne(df.A.shift()).cumsum()-1
df
When we print df
value it will be as follows.
Index A B C
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
Explanation of solution: Let's break above solution into parts for understanding purposes.
1st step: Compare df's A column by shifting its value down to itself as follows.
i.ne(df.A.shift())
Output we will get is:
0 True
1 False
2 False
3 True
4 False
5 True
2nd step: Use of cumsum()
function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum()
function and its value will be increased.
i.ne(df.A.shift()).cumsum()-1
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int32
3rd step: Save command's value into df['C']
which will create a new column named C
in df
.
1
Nice method ve++ for you
– W-B
36 mins ago
@W-B, thank you for encouragement sir, ++ve for your unique style already :)
– RavinderSingh13
34 mins ago
add a comment |
up vote
2
down vote
up vote
2
down vote
One more method of doing so could be.
df['C'] = i.ne(df.A.shift()).cumsum()-1
df
When we print df
value it will be as follows.
Index A B C
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
Explanation of solution: Let's break above solution into parts for understanding purposes.
1st step: Compare df's A column by shifting its value down to itself as follows.
i.ne(df.A.shift())
Output we will get is:
0 True
1 False
2 False
3 True
4 False
5 True
2nd step: Use of cumsum()
function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum()
function and its value will be increased.
i.ne(df.A.shift()).cumsum()-1
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int32
3rd step: Save command's value into df['C']
which will create a new column named C
in df
.
One more method of doing so could be.
df['C'] = i.ne(df.A.shift()).cumsum()-1
df
When we print df
value it will be as follows.
Index A B C
0 0 foo 3 0
1 1 foo 2 0
2 2 foo 5 0
3 3 bar 3 1
4 4 bar 4 1
5 5 baz 5 2
Explanation of solution: Let's break above solution into parts for understanding purposes.
1st step: Compare df's A column by shifting its value down to itself as follows.
i.ne(df.A.shift())
Output we will get is:
0 True
1 False
2 False
3 True
4 False
5 True
2nd step: Use of cumsum()
function, so wherever TRUE value is coming(which will come when a match of A column and its shift is NOT found) it will call cumsum()
function and its value will be increased.
i.ne(df.A.shift()).cumsum()-1
0 0
1 0
2 0
3 1
4 1
5 2
Name: A, dtype: int32
3rd step: Save command's value into df['C']
which will create a new column named C
in df
.
edited 23 mins ago
answered 42 mins ago
RavinderSingh13
25k41437
25k41437
1
Nice method ve++ for you
– W-B
36 mins ago
@W-B, thank you for encouragement sir, ++ve for your unique style already :)
– RavinderSingh13
34 mins ago
add a comment |
1
Nice method ve++ for you
– W-B
36 mins ago
@W-B, thank you for encouragement sir, ++ve for your unique style already :)
– RavinderSingh13
34 mins ago
1
1
Nice method ve++ for you
– W-B
36 mins ago
Nice method ve++ for you
– W-B
36 mins ago
@W-B, thank you for encouragement sir, ++ve for your unique style already :)
– RavinderSingh13
34 mins ago
@W-B, thank you for encouragement sir, ++ve for your unique style already :)
– RavinderSingh13
34 mins ago
add a comment |
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