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6

I have a dataframe like below:

A B C
1 1 1
2 0 1
3 0 0
4 1 0
5 0 1
6 0 0
7 1 0

I want the number of occurence of zeroes from df['B'] under the following condition:

if(df['B']<df['C']):
 #count number of zeroes in df['B'] until it sees 1.

expected output:

A B C output
1 1 1 Nan
2 0 1 1
3 0 0 Nan
4 1 0 Nan
5 0 1 1
6 0 1 0
7 1 0 Nan

I dont know how to formulate the count part. Any help is really appreciated

python pandas dataframe

share|improve this question

edited 8 hours ago

Massifox

54211 silver badge1313 bronze badges

asked 9 hours ago

hakuna_codehakuna_code

15188 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Me too, what does until it sees 1 mean?
  
  – Joe
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  until the first occurence of '1' in B
  
  – hakuna_code
  9 hours ago
  

  
  
  
  

add a comment
 | 

6

I have a dataframe like below:

A B C
1 1 1
2 0 1
3 0 0
4 1 0
5 0 1
6 0 0
7 1 0

I want the number of occurence of zeroes from df['B'] under the following condition:

if(df['B']<df['C']):
 #count number of zeroes in df['B'] until it sees 1.

expected output:

A B C output
1 1 1 Nan
2 0 1 1
3 0 0 Nan
4 1 0 Nan
5 0 1 1
6 0 1 0
7 1 0 Nan

I dont know how to formulate the count part. Any help is really appreciated

python pandas dataframe

share|improve this question

edited 8 hours ago

Massifox

54211 silver badge1313 bronze badges

asked 9 hours ago

hakuna_codehakuna_code

15188 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Me too, what does until it sees 1 mean?
  
  – Joe
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  until the first occurence of '1' in B
  
  – hakuna_code
  9 hours ago
  

  
  
  
  

add a comment
 | 

I have a dataframe like below:

A B C
1 1 1
2 0 1
3 0 0
4 1 0
5 0 1
6 0 0
7 1 0

I want the number of occurence of zeroes from df['B'] under the following condition:

if(df['B']<df['C']):
 #count number of zeroes in df['B'] until it sees 1.

expected output:

A B C output
1 1 1 Nan
2 0 1 1
3 0 0 Nan
4 1 0 Nan
5 0 1 1
6 0 1 0
7 1 0 Nan

I dont know how to formulate the count part. Any help is really appreciated

python pandas dataframe

share|improve this question

edited 8 hours ago

Massifox

54211 silver badge1313 bronze badges

asked 9 hours ago

hakuna_codehakuna_code

15188 bronze badges

A B C
1 1 1
2 0 1
3 0 0
4 1 0
5 0 1
6 0 0
7 1 0

if(df['B']<df['C']):
 #count number of zeroes in df['B'] until it sees 1.

A B C output
1 1 1 Nan
2 0 1 1
3 0 0 Nan
4 1 0 Nan
5 0 1 1
6 0 1 0
7 1 0 Nan

share|improve this question

edited 8 hours ago

Massifox

54211 silver badge1313 bronze badges

asked 9 hours ago

hakuna_codehakuna_code

15188 bronze badges

share|improve this question

edited 8 hours ago

Massifox

54211 silver badge1313 bronze badges

asked 9 hours ago

hakuna_codehakuna_code

15188 bronze badges

edited 8 hours ago

Massifox

54211 silver badge1313 bronze badges

edited 8 hours ago

Massifox

54211 silver badge1313 bronze badges

asked 9 hours ago

hakuna_codehakuna_code

15188 bronze badges

asked 9 hours ago

hakuna_codehakuna_code

15188 bronze badges

3 Answers
3

active

oldest

votes

5

IIUC one approach would be using a custom grouper and aggregating with groupby.cumcount:

c1 = df.B.lt(df.C)
g = df.B.eq(1).cumsum()
df['out'] = c1.groupby(g).cumcount(ascending=False).shift().where(c1).sub(1)

---

print(df)

 A B C out
0 1 1 1 NaN
1 2 0 1 1.0
2 3 0 0 NaN
3 4 1 0 NaN
4 5 0 1 1.0
5 6 0 1 0.0
6 7 1 0 NaN

share|improve this answer

answered 9 hours ago

yatuyatu

32.6k66 gold badges2626 silver badges5858 bronze badges

add a comment
 | 

6

Using some masking and a groupby on your reversed series. This assumes binary data (only 0 and 1)

---

m = df['B'][::-1].eq(0)
d = m.groupby(m.ne(m.shift()).cumsum()).cumsum().sub(1)
d[::-1].where(df['B'] < df['C'])

0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
Name: B, dtype: float64

And a fast numpy based approach

def zero_until_one(a, b):
 n = a.shape[0] 
 x = np.flatnonzero(a < b)
 y = np.flatnonzero(a == 1) 
 d = np.searchsorted(y, x)
 r = y[d] - x - 1
 out = np.full(n, np.nan)
 out[x] = r 
 return out

zero_until_one(df['B'], df['C'])

array([nan, 1., nan, nan, 1., 0., nan])

Performance

df = pd.concat([df]*10_000)

%timeit chris1(df)
19.3 ms ± 348 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit yatu(df)
12.8 ms ± 54.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit zero_until_one(df['B'], df['C'])
2.32 ms ± 31.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

share|improve this answer

edited 8 hours ago

answered 9 hours ago

user3483203user3483203

39k88 gold badges3232 silver badges6363 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Great idea for numpy function , Just guess numba may faster
  
  – WeNYoBen
  8 hours ago
  

  
  
  
  

add a comment
 | 

1

Let us push into one-line

df.groupby(df.B.iloc[::-1].cumsum()).cumcount(ascending=False).shift(-1).where(df.B<df.C)
Out[80]: 
0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
dtype: float64

share|improve this answer

answered 8 hours ago

WeNYoBenWeNYoBen

158k88 gold badges5454 silver badges8686 bronze badges

add a comment
 | 

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5

IIUC one approach would be using a custom grouper and aggregating with groupby.cumcount:

c1 = df.B.lt(df.C)
g = df.B.eq(1).cumsum()
df['out'] = c1.groupby(g).cumcount(ascending=False).shift().where(c1).sub(1)

---

print(df)

 A B C out
0 1 1 1 NaN
1 2 0 1 1.0
2 3 0 0 NaN
3 4 1 0 NaN
4 5 0 1 1.0
5 6 0 1 0.0
6 7 1 0 NaN

share|improve this answer

answered 9 hours ago

yatuyatu

32.6k66 gold badges2626 silver badges5858 bronze badges

add a comment
 | 

5

IIUC one approach would be using a custom grouper and aggregating with groupby.cumcount:

c1 = df.B.lt(df.C)
g = df.B.eq(1).cumsum()
df['out'] = c1.groupby(g).cumcount(ascending=False).shift().where(c1).sub(1)

---

print(df)

 A B C out
0 1 1 1 NaN
1 2 0 1 1.0
2 3 0 0 NaN
3 4 1 0 NaN
4 5 0 1 1.0
5 6 0 1 0.0
6 7 1 0 NaN

share|improve this answer

answered 9 hours ago

yatuyatu

32.6k66 gold badges2626 silver badges5858 bronze badges

add a comment
 | 

IIUC one approach would be using a custom grouper and aggregating with groupby.cumcount:

c1 = df.B.lt(df.C)
g = df.B.eq(1).cumsum()
df['out'] = c1.groupby(g).cumcount(ascending=False).shift().where(c1).sub(1)

---

print(df)

 A B C out
0 1 1 1 NaN
1 2 0 1 1.0
2 3 0 0 NaN
3 4 1 0 NaN
4 5 0 1 1.0
5 6 0 1 0.0
6 7 1 0 NaN

share|improve this answer

answered 9 hours ago

yatuyatu

32.6k66 gold badges2626 silver badges5858 bronze badges

c1 = df.B.lt(df.C)
g = df.B.eq(1).cumsum()
df['out'] = c1.groupby(g).cumcount(ascending=False).shift().where(c1).sub(1)

print(df)

 A B C out
0 1 1 1 NaN
1 2 0 1 1.0
2 3 0 0 NaN
3 4 1 0 NaN
4 5 0 1 1.0
5 6 0 1 0.0
6 7 1 0 NaN

share|improve this answer

answered 9 hours ago

yatuyatu

32.6k66 gold badges2626 silver badges5858 bronze badges

answered 9 hours ago

yatuyatu

32.6k66 gold badges2626 silver badges5858 bronze badges

answered 9 hours ago

yatuyatu

32.6k66 gold badges2626 silver badges5858 bronze badges

6

Using some masking and a groupby on your reversed series. This assumes binary data (only 0 and 1)

---

m = df['B'][::-1].eq(0)
d = m.groupby(m.ne(m.shift()).cumsum()).cumsum().sub(1)
d[::-1].where(df['B'] < df['C'])

0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
Name: B, dtype: float64

And a fast numpy based approach

def zero_until_one(a, b):
 n = a.shape[0] 
 x = np.flatnonzero(a < b)
 y = np.flatnonzero(a == 1) 
 d = np.searchsorted(y, x)
 r = y[d] - x - 1
 out = np.full(n, np.nan)
 out[x] = r 
 return out

zero_until_one(df['B'], df['C'])

array([nan, 1., nan, nan, 1., 0., nan])

Performance

df = pd.concat([df]*10_000)

%timeit chris1(df)
19.3 ms ± 348 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit yatu(df)
12.8 ms ± 54.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit zero_until_one(df['B'], df['C'])
2.32 ms ± 31.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

share|improve this answer

edited 8 hours ago

answered 9 hours ago

user3483203user3483203

39k88 gold badges3232 silver badges6363 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Great idea for numpy function , Just guess numba may faster
  
  – WeNYoBen
  8 hours ago
  

  
  
  
  

add a comment
 | 

6

Using some masking and a groupby on your reversed series. This assumes binary data (only 0 and 1)

---

m = df['B'][::-1].eq(0)
d = m.groupby(m.ne(m.shift()).cumsum()).cumsum().sub(1)
d[::-1].where(df['B'] < df['C'])

0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
Name: B, dtype: float64

And a fast numpy based approach

def zero_until_one(a, b):
 n = a.shape[0] 
 x = np.flatnonzero(a < b)
 y = np.flatnonzero(a == 1) 
 d = np.searchsorted(y, x)
 r = y[d] - x - 1
 out = np.full(n, np.nan)
 out[x] = r 
 return out

zero_until_one(df['B'], df['C'])

array([nan, 1., nan, nan, 1., 0., nan])

Performance

df = pd.concat([df]*10_000)

%timeit chris1(df)
19.3 ms ± 348 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit yatu(df)
12.8 ms ± 54.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit zero_until_one(df['B'], df['C'])
2.32 ms ± 31.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

share|improve this answer

edited 8 hours ago

answered 9 hours ago

user3483203user3483203

39k88 gold badges3232 silver badges6363 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Great idea for numpy function , Just guess numba may faster
  
  – WeNYoBen
  8 hours ago
  

  
  
  
  

add a comment
 | 

Using some masking and a groupby on your reversed series. This assumes binary data (only 0 and 1)

---

m = df['B'][::-1].eq(0)
d = m.groupby(m.ne(m.shift()).cumsum()).cumsum().sub(1)
d[::-1].where(df['B'] < df['C'])

0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
Name: B, dtype: float64

And a fast numpy based approach

def zero_until_one(a, b):
 n = a.shape[0] 
 x = np.flatnonzero(a < b)
 y = np.flatnonzero(a == 1) 
 d = np.searchsorted(y, x)
 r = y[d] - x - 1
 out = np.full(n, np.nan)
 out[x] = r 
 return out

zero_until_one(df['B'], df['C'])

array([nan, 1., nan, nan, 1., 0., nan])

Performance

df = pd.concat([df]*10_000)

%timeit chris1(df)
19.3 ms ± 348 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit yatu(df)
12.8 ms ± 54.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit zero_until_one(df['B'], df['C'])
2.32 ms ± 31.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

share|improve this answer

edited 8 hours ago

answered 9 hours ago

user3483203user3483203

39k88 gold badges3232 silver badges6363 bronze badges

m = df['B'][::-1].eq(0)
d = m.groupby(m.ne(m.shift()).cumsum()).cumsum().sub(1)
d[::-1].where(df['B'] < df['C'])

0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
Name: B, dtype: float64

def zero_until_one(a, b):
 n = a.shape[0] 
 x = np.flatnonzero(a < b)
 y = np.flatnonzero(a == 1) 
 d = np.searchsorted(y, x)
 r = y[d] - x - 1
 out = np.full(n, np.nan)
 out[x] = r 
 return out

zero_until_one(df['B'], df['C'])

array([nan, 1., nan, nan, 1., 0., nan])

df = pd.concat([df]*10_000)

%timeit chris1(df)
19.3 ms ± 348 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit yatu(df)
12.8 ms ± 54.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit zero_until_one(df['B'], df['C'])
2.32 ms ± 31.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

share|improve this answer

edited 8 hours ago

answered 9 hours ago

user3483203user3483203

39k88 gold badges3232 silver badges6363 bronze badges

answered 9 hours ago

user3483203user3483203

39k88 gold badges3232 silver badges6363 bronze badges

answered 9 hours ago

user3483203user3483203

39k88 gold badges3232 silver badges6363 bronze badges

1

Let us push into one-line

df.groupby(df.B.iloc[::-1].cumsum()).cumcount(ascending=False).shift(-1).where(df.B<df.C)
Out[80]: 
0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
dtype: float64

share|improve this answer

answered 8 hours ago

WeNYoBenWeNYoBen

158k88 gold badges5454 silver badges8686 bronze badges

add a comment
 | 

1

Let us push into one-line

df.groupby(df.B.iloc[::-1].cumsum()).cumcount(ascending=False).shift(-1).where(df.B<df.C)
Out[80]: 
0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
dtype: float64

share|improve this answer

answered 8 hours ago

WeNYoBenWeNYoBen

158k88 gold badges5454 silver badges8686 bronze badges

add a comment
 | 

Let us push into one-line

df.groupby(df.B.iloc[::-1].cumsum()).cumcount(ascending=False).shift(-1).where(df.B<df.C)
Out[80]: 
0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
dtype: float64

share|improve this answer

answered 8 hours ago

WeNYoBenWeNYoBen

158k88 gold badges5454 silver badges8686 bronze badges

df.groupby(df.B.iloc[::-1].cumsum()).cumcount(ascending=False).shift(-1).where(df.B<df.C)
Out[80]: 
0 NaN
1 1.0
2 NaN
3 NaN
4 1.0
5 0.0
6 NaN
dtype: float64

share|improve this answer

answered 8 hours ago

WeNYoBenWeNYoBen

158k88 gold badges5454 silver badges8686 bronze badges

answered 8 hours ago

WeNYoBenWeNYoBen

158k88 gold badges5454 silver badges8686 bronze badges

answered 8 hours ago

WeNYoBenWeNYoBen

158k88 gold badges5454 silver badges8686 bronze badges