Mixing
Pandas: conditional counting when multiple conditions are met
2
I have a dataframe as follows:
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.000000 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.000000 0 0
2 2018-03-01 00:00:02 +0000 50.120 9.000000 0 0
3 2018-03-01 00:00:03 +0000 50.112 9.000000 0 0
4 2018-03-01 00:00:04 +0000 50.102 9.000000 0 0
5 2018-03-01 00:00:05 +0000 50.097 9.000000 0 0
6 2018-03-01 00:00:06 +0000 11.095 9.000000 0 0
7 2018-03-01 00:00:07 +0000 11.095 9.000000 0 0
8 2018-03-01 00:00:08 +0000 11.092 9.000000 0 0
9 2018-03-01 00:00:09 +0000 11.095 9.000000 0 0
10 2018-03-01 00:00:10 +0000 11.097 5.000000 0 0
11 2018-03-01 00:00:11 +0000 11.097 5.000000 0 0
12 2018-03-01 00:00:12 +0000 11.097 5.000000 0 0
13 2018-03-01 00:00:13 +0000 50.100 5.000000 0 0
14 2018-03-01 00:00:14 +0000 50.102 5.000000 0 0
15 2018-03-01 00:00:15 +0000 50.105 5.000000 0 0
16 2018-03-01 00:00:16 +0000 50.102 5.000000 0 0
17 2018-03-01 00:00:17 +0000 50.102 5.000000 0 0
A and B are two Counters that work like this:
if((f>=50) or (f<50 & C<8)) then A increase by 1
if f<50 and C>8 then B increase by 1
the expected outcome should be like:
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.000000 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.000000 1 0
2 2018-03-01 00:00:02 +0000 50.120 9.000000 2 0
3 2018-03-01 00:00:03 +0000 50.112 9.000000 3 0
4 2018-03-01 00:00:04 +0000 50.102 9.000000 4 0
5 2018-03-01 00:00:05 +0000 50.097 9.000000 5 0
6 2018-03-01 00:00:06 +0000 11.095 9.000000 5 1
7 2018-03-01 00:00:07 +0000 11.095 9.000000 5 2
8 2018-03-01 00:00:08 +0000 11.092 9.000000 5 3
9 2018-03-01 00:00:09 +0000 11.095 9.000000 5 4
10 2018-03-01 00:00:10 +0000 11.097 5.000000 6 4
11 2018-03-01 00:00:11 +0000 11.097 5.000000 7 4
12 2018-03-01 00:00:12 +0000 11.097 5.000000 8 4
13 2018-03-01 00:00:13 +0000 50.100 5.000000 9 4
14 2018-03-01 00:00:14 +0000 50.102 5.000000 10 4
15 2018-03-01 00:00:15 +0000 50.105 5.000000 11 4
16 2018-03-01 00:00:16 +0000 50.102 5.000000 12 4
17 2018-03-01 00:00:17 +0000 50.102 5.000000 13 4
Please notice that when a A increases B keeps its value, and the other way around. They do not reset. Any idea about that?
Thank you in advance!
python pandas
asked Nov 20 '18 at 15:00
Luca91Luca91
1808
add a comment |
2
I have a dataframe as follows:
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.000000 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.000000 0 0
2 2018-03-01 00:00:02 +0000 50.120 9.000000 0 0
3 2018-03-01 00:00:03 +0000 50.112 9.000000 0 0
4 2018-03-01 00:00:04 +0000 50.102 9.000000 0 0
5 2018-03-01 00:00:05 +0000 50.097 9.000000 0 0
6 2018-03-01 00:00:06 +0000 11.095 9.000000 0 0
7 2018-03-01 00:00:07 +0000 11.095 9.000000 0 0
8 2018-03-01 00:00:08 +0000 11.092 9.000000 0 0
9 2018-03-01 00:00:09 +0000 11.095 9.000000 0 0
10 2018-03-01 00:00:10 +0000 11.097 5.000000 0 0
11 2018-03-01 00:00:11 +0000 11.097 5.000000 0 0
12 2018-03-01 00:00:12 +0000 11.097 5.000000 0 0
13 2018-03-01 00:00:13 +0000 50.100 5.000000 0 0
14 2018-03-01 00:00:14 +0000 50.102 5.000000 0 0
15 2018-03-01 00:00:15 +0000 50.105 5.000000 0 0
16 2018-03-01 00:00:16 +0000 50.102 5.000000 0 0
17 2018-03-01 00:00:17 +0000 50.102 5.000000 0 0
A and B are two Counters that work like this:
if((f>=50) or (f<50 & C<8)) then A increase by 1
if f<50 and C>8 then B increase by 1
the expected outcome should be like:
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.000000 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.000000 1 0
2 2018-03-01 00:00:02 +0000 50.120 9.000000 2 0
3 2018-03-01 00:00:03 +0000 50.112 9.000000 3 0
4 2018-03-01 00:00:04 +0000 50.102 9.000000 4 0
5 2018-03-01 00:00:05 +0000 50.097 9.000000 5 0
6 2018-03-01 00:00:06 +0000 11.095 9.000000 5 1
7 2018-03-01 00:00:07 +0000 11.095 9.000000 5 2
8 2018-03-01 00:00:08 +0000 11.092 9.000000 5 3
9 2018-03-01 00:00:09 +0000 11.095 9.000000 5 4
10 2018-03-01 00:00:10 +0000 11.097 5.000000 6 4
11 2018-03-01 00:00:11 +0000 11.097 5.000000 7 4
12 2018-03-01 00:00:12 +0000 11.097 5.000000 8 4
13 2018-03-01 00:00:13 +0000 50.100 5.000000 9 4
14 2018-03-01 00:00:14 +0000 50.102 5.000000 10 4
15 2018-03-01 00:00:15 +0000 50.105 5.000000 11 4
16 2018-03-01 00:00:16 +0000 50.102 5.000000 12 4
17 2018-03-01 00:00:17 +0000 50.102 5.000000 13 4
Please notice that when a A increases B keeps its value, and the other way around. They do not reset. Any idea about that?
Thank you in advance!
python pandas
asked Nov 20 '18 at 15:00
Luca91Luca91
1808
add a comment |
2
2
2
I have a dataframe as follows:
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.000000 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.000000 0 0
2 2018-03-01 00:00:02 +0000 50.120 9.000000 0 0
3 2018-03-01 00:00:03 +0000 50.112 9.000000 0 0
4 2018-03-01 00:00:04 +0000 50.102 9.000000 0 0
5 2018-03-01 00:00:05 +0000 50.097 9.000000 0 0
6 2018-03-01 00:00:06 +0000 11.095 9.000000 0 0
7 2018-03-01 00:00:07 +0000 11.095 9.000000 0 0
8 2018-03-01 00:00:08 +0000 11.092 9.000000 0 0
9 2018-03-01 00:00:09 +0000 11.095 9.000000 0 0
10 2018-03-01 00:00:10 +0000 11.097 5.000000 0 0
11 2018-03-01 00:00:11 +0000 11.097 5.000000 0 0
12 2018-03-01 00:00:12 +0000 11.097 5.000000 0 0
13 2018-03-01 00:00:13 +0000 50.100 5.000000 0 0
14 2018-03-01 00:00:14 +0000 50.102 5.000000 0 0
15 2018-03-01 00:00:15 +0000 50.105 5.000000 0 0
16 2018-03-01 00:00:16 +0000 50.102 5.000000 0 0
17 2018-03-01 00:00:17 +0000 50.102 5.000000 0 0
A and B are two Counters that work like this:
if((f>=50) or (f<50 & C<8)) then A increase by 1
if f<50 and C>8 then B increase by 1
the expected outcome should be like:
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.000000 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.000000 1 0
2 2018-03-01 00:00:02 +0000 50.120 9.000000 2 0
3 2018-03-01 00:00:03 +0000 50.112 9.000000 3 0
4 2018-03-01 00:00:04 +0000 50.102 9.000000 4 0
5 2018-03-01 00:00:05 +0000 50.097 9.000000 5 0
6 2018-03-01 00:00:06 +0000 11.095 9.000000 5 1
7 2018-03-01 00:00:07 +0000 11.095 9.000000 5 2
8 2018-03-01 00:00:08 +0000 11.092 9.000000 5 3
9 2018-03-01 00:00:09 +0000 11.095 9.000000 5 4
10 2018-03-01 00:00:10 +0000 11.097 5.000000 6 4
11 2018-03-01 00:00:11 +0000 11.097 5.000000 7 4
12 2018-03-01 00:00:12 +0000 11.097 5.000000 8 4
13 2018-03-01 00:00:13 +0000 50.100 5.000000 9 4
14 2018-03-01 00:00:14 +0000 50.102 5.000000 10 4
15 2018-03-01 00:00:15 +0000 50.105 5.000000 11 4
16 2018-03-01 00:00:16 +0000 50.102 5.000000 12 4
17 2018-03-01 00:00:17 +0000 50.102 5.000000 13 4
Please notice that when a A increases B keeps its value, and the other way around. They do not reset. Any idea about that?
Thank you in advance!
python pandas
asked Nov 20 '18 at 15:00
Luca91Luca91
1808
I have a dataframe as follows:
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.000000 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.000000 0 0
2 2018-03-01 00:00:02 +0000 50.120 9.000000 0 0
3 2018-03-01 00:00:03 +0000 50.112 9.000000 0 0
4 2018-03-01 00:00:04 +0000 50.102 9.000000 0 0
5 2018-03-01 00:00:05 +0000 50.097 9.000000 0 0
6 2018-03-01 00:00:06 +0000 11.095 9.000000 0 0
7 2018-03-01 00:00:07 +0000 11.095 9.000000 0 0
8 2018-03-01 00:00:08 +0000 11.092 9.000000 0 0
9 2018-03-01 00:00:09 +0000 11.095 9.000000 0 0
10 2018-03-01 00:00:10 +0000 11.097 5.000000 0 0
11 2018-03-01 00:00:11 +0000 11.097 5.000000 0 0
12 2018-03-01 00:00:12 +0000 11.097 5.000000 0 0
13 2018-03-01 00:00:13 +0000 50.100 5.000000 0 0
14 2018-03-01 00:00:14 +0000 50.102 5.000000 0 0
15 2018-03-01 00:00:15 +0000 50.105 5.000000 0 0
16 2018-03-01 00:00:16 +0000 50.102 5.000000 0 0
17 2018-03-01 00:00:17 +0000 50.102 5.000000 0 0
A and B are two Counters that work like this:
if((f>=50) or (f<50 & C<8)) then A increase by 1
if f<50 and C>8 then B increase by 1
the expected outcome should be like:
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.000000 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.000000 1 0
2 2018-03-01 00:00:02 +0000 50.120 9.000000 2 0
3 2018-03-01 00:00:03 +0000 50.112 9.000000 3 0
4 2018-03-01 00:00:04 +0000 50.102 9.000000 4 0
5 2018-03-01 00:00:05 +0000 50.097 9.000000 5 0
6 2018-03-01 00:00:06 +0000 11.095 9.000000 5 1
7 2018-03-01 00:00:07 +0000 11.095 9.000000 5 2
8 2018-03-01 00:00:08 +0000 11.092 9.000000 5 3
9 2018-03-01 00:00:09 +0000 11.095 9.000000 5 4
10 2018-03-01 00:00:10 +0000 11.097 5.000000 6 4
11 2018-03-01 00:00:11 +0000 11.097 5.000000 7 4
12 2018-03-01 00:00:12 +0000 11.097 5.000000 8 4
13 2018-03-01 00:00:13 +0000 50.100 5.000000 9 4
14 2018-03-01 00:00:14 +0000 50.102 5.000000 10 4
15 2018-03-01 00:00:15 +0000 50.105 5.000000 11 4
16 2018-03-01 00:00:16 +0000 50.102 5.000000 12 4
17 2018-03-01 00:00:17 +0000 50.102 5.000000 13 4
Please notice that when a A increases B keeps its value, and the other way around. They do not reset. Any idea about that?
Thank you in advance!
python pandas
python pandas
asked Nov 20 '18 at 15:00
Luca91Luca91
1808
asked Nov 20 '18 at 15:00
Luca91Luca91
1808
asked Nov 20 '18 at 15:00
Luca91Luca91
1808
asked Nov 20 '18 at 15:00
Luca91Luca91
1808
asked Nov 20 '18 at 15:00
Luca91Luca91
1808
1808
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
5
For me working nice subtracting 1 with sub and for removing possible -1 in first rows add clip_lower:
m1 = (df.f >=50) | ((df.f<50) & (df.C<8))
m2 = (df.f<50) & (df.C>8)
df['A'] = m1.cumsum().sub(1).clip_lower(0)
df['B'] = m2.cumsum().sub(1).clip_lower(0)
answered Nov 20 '18 at 15:04
jezraeljezrael
329k23270349
3
@Lucas91 boolean series withcumsumis the trick.
– Scott Boston
Nov 20 '18 at 15:07
@jezrael, exccellent answer, may I know why there is a change in your output, and the expected output in question. there is no '4' in your df['B']
– pyd
Nov 20 '18 at 16:21
add a comment |
5
Assumptions
df.C > 8was meant to bedf.C >= 8because that would be the compliment todf.C < 8
(df.f < 50) & (df.C < 8)isn't necessary because of theorstatement anddf.f >= 50on the other side of it.- Column
'A'starting with0seems to be a weird thing that needs special handling. It would be cleaner to assume that it begins with zero and starts incrementing at the firstTrue
In line with assign
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df.assign(A=c.cumsum(), B=(~c).cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 1 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 2 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 3 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 4 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 5 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 6 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 6 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 6 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 6 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 6 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 7 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 8 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 9 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 10 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 11 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 12 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 13 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 14 4
In place
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df[['A', 'B']] = np.column_stack([c, ~c]).cumsum(0)
df
Reduced
c = (df.f.values >= 50) | (df.C.values < 8)
df.assign(A=c.cumsum(), B=(~c).cumsum())
With special handling
a = df.f.values >= 50
b = df.C.values < 8
c0 = a | b
c1 = ~c0
c0[0] = False
c1[0] = False
df.assign(A=c0.cumsum(), B=c1.cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 1 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 2 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 3 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 4 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 5 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 5 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 5 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 5 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 5 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 6 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 7 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 8 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 9 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 10 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 11 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 12 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 13 4
answered Nov 20 '18 at 15:30
piRSquaredpiRSquared
154k22146288
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
For me working nice subtracting 1 with sub and for removing possible -1 in first rows add clip_lower:
m1 = (df.f >=50) | ((df.f<50) & (df.C<8))
m2 = (df.f<50) & (df.C>8)
df['A'] = m1.cumsum().sub(1).clip_lower(0)
df['B'] = m2.cumsum().sub(1).clip_lower(0)
answered Nov 20 '18 at 15:04
jezraeljezrael
329k23270349
3
@Lucas91 boolean series withcumsumis the trick.
– Scott Boston
Nov 20 '18 at 15:07
@jezrael, exccellent answer, may I know why there is a change in your output, and the expected output in question. there is no '4' in your df['B']
– pyd
Nov 20 '18 at 16:21
add a comment |
5
For me working nice subtracting 1 with sub and for removing possible -1 in first rows add clip_lower:
m1 = (df.f >=50) | ((df.f<50) & (df.C<8))
m2 = (df.f<50) & (df.C>8)
df['A'] = m1.cumsum().sub(1).clip_lower(0)
df['B'] = m2.cumsum().sub(1).clip_lower(0)
answered Nov 20 '18 at 15:04
jezraeljezrael
329k23270349
3
@Lucas91 boolean series withcumsumis the trick.
– Scott Boston
Nov 20 '18 at 15:07
@jezrael, exccellent answer, may I know why there is a change in your output, and the expected output in question. there is no '4' in your df['B']
– pyd
Nov 20 '18 at 16:21
add a comment |
5
5
5
For me working nice subtracting 1 with sub and for removing possible -1 in first rows add clip_lower:
m1 = (df.f >=50) | ((df.f<50) & (df.C<8))
m2 = (df.f<50) & (df.C>8)
df['A'] = m1.cumsum().sub(1).clip_lower(0)
df['B'] = m2.cumsum().sub(1).clip_lower(0)
answered Nov 20 '18 at 15:04
jezraeljezrael
329k23270349
For me working nice subtracting 1 with sub and for removing possible -1 in first rows add clip_lower:
m1 = (df.f >=50) | ((df.f<50) & (df.C<8))
m2 = (df.f<50) & (df.C>8)
df['A'] = m1.cumsum().sub(1).clip_lower(0)
df['B'] = m2.cumsum().sub(1).clip_lower(0)
answered Nov 20 '18 at 15:04
jezraeljezrael
329k23270349
edited Nov 20 '18 at 15:11
answered Nov 20 '18 at 15:04
jezraeljezrael
329k23270349
answered Nov 20 '18 at 15:04
jezraeljezrael
329k23270349
answered Nov 20 '18 at 15:04
jezraeljezrael
329k23270349
329k23270349
3
@Lucas91 boolean series withcumsumis the trick.
– Scott Boston
Nov 20 '18 at 15:07
@jezrael, exccellent answer, may I know why there is a change in your output, and the expected output in question. there is no '4' in your df['B']
– pyd
Nov 20 '18 at 16:21
add a comment |
3
@Lucas91 boolean series withcumsumis the trick.
– Scott Boston
Nov 20 '18 at 15:07
@jezrael, exccellent answer, may I know why there is a change in your output, and the expected output in question. there is no '4' in your df['B']
– pyd
Nov 20 '18 at 16:21
3
3
@Lucas91 boolean series with
cumsum is the trick.– Scott Boston
Nov 20 '18 at 15:07
@Lucas91 boolean series with
cumsum is the trick.– Scott Boston
Nov 20 '18 at 15:07
@jezrael, exccellent answer, may I know why there is a change in your output, and the expected output in question. there is no '4' in your df['B']
– pyd
Nov 20 '18 at 16:21
@jezrael, exccellent answer, may I know why there is a change in your output, and the expected output in question. there is no '4' in your df['B']
– pyd
Nov 20 '18 at 16:21
add a comment |
5
Assumptions
df.C > 8was meant to bedf.C >= 8because that would be the compliment todf.C < 8
(df.f < 50) & (df.C < 8)isn't necessary because of theorstatement anddf.f >= 50on the other side of it.- Column
'A'starting with0seems to be a weird thing that needs special handling. It would be cleaner to assume that it begins with zero and starts incrementing at the firstTrue
In line with assign
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df.assign(A=c.cumsum(), B=(~c).cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 1 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 2 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 3 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 4 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 5 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 6 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 6 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 6 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 6 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 6 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 7 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 8 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 9 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 10 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 11 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 12 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 13 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 14 4
In place
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df[['A', 'B']] = np.column_stack([c, ~c]).cumsum(0)
df
Reduced
c = (df.f.values >= 50) | (df.C.values < 8)
df.assign(A=c.cumsum(), B=(~c).cumsum())
With special handling
a = df.f.values >= 50
b = df.C.values < 8
c0 = a | b
c1 = ~c0
c0[0] = False
c1[0] = False
df.assign(A=c0.cumsum(), B=c1.cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 1 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 2 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 3 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 4 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 5 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 5 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 5 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 5 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 5 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 6 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 7 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 8 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 9 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 10 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 11 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 12 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 13 4
answered Nov 20 '18 at 15:30
piRSquaredpiRSquared
154k22146288
add a comment |
5
Assumptions
df.C > 8was meant to bedf.C >= 8because that would be the compliment todf.C < 8
(df.f < 50) & (df.C < 8)isn't necessary because of theorstatement anddf.f >= 50on the other side of it.- Column
'A'starting with0seems to be a weird thing that needs special handling. It would be cleaner to assume that it begins with zero and starts incrementing at the firstTrue
In line with assign
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df.assign(A=c.cumsum(), B=(~c).cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 1 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 2 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 3 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 4 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 5 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 6 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 6 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 6 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 6 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 6 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 7 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 8 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 9 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 10 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 11 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 12 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 13 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 14 4
In place
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df[['A', 'B']] = np.column_stack([c, ~c]).cumsum(0)
df
Reduced
c = (df.f.values >= 50) | (df.C.values < 8)
df.assign(A=c.cumsum(), B=(~c).cumsum())
With special handling
a = df.f.values >= 50
b = df.C.values < 8
c0 = a | b
c1 = ~c0
c0[0] = False
c1[0] = False
df.assign(A=c0.cumsum(), B=c1.cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 1 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 2 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 3 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 4 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 5 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 5 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 5 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 5 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 5 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 6 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 7 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 8 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 9 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 10 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 11 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 12 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 13 4
answered Nov 20 '18 at 15:30
piRSquaredpiRSquared
154k22146288
add a comment |
5
5
5
Assumptions
df.C > 8was meant to bedf.C >= 8because that would be the compliment todf.C < 8
(df.f < 50) & (df.C < 8)isn't necessary because of theorstatement anddf.f >= 50on the other side of it.- Column
'A'starting with0seems to be a weird thing that needs special handling. It would be cleaner to assume that it begins with zero and starts incrementing at the firstTrue
In line with assign
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df.assign(A=c.cumsum(), B=(~c).cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 1 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 2 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 3 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 4 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 5 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 6 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 6 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 6 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 6 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 6 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 7 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 8 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 9 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 10 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 11 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 12 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 13 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 14 4
In place
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df[['A', 'B']] = np.column_stack([c, ~c]).cumsum(0)
df
Reduced
c = (df.f.values >= 50) | (df.C.values < 8)
df.assign(A=c.cumsum(), B=(~c).cumsum())
With special handling
a = df.f.values >= 50
b = df.C.values < 8
c0 = a | b
c1 = ~c0
c0[0] = False
c1[0] = False
df.assign(A=c0.cumsum(), B=c1.cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 1 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 2 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 3 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 4 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 5 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 5 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 5 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 5 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 5 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 6 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 7 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 8 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 9 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 10 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 11 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 12 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 13 4
answered Nov 20 '18 at 15:30
piRSquaredpiRSquared
154k22146288
Assumptions
df.C > 8was meant to bedf.C >= 8because that would be the compliment todf.C < 8
(df.f < 50) & (df.C < 8)isn't necessary because of theorstatement anddf.f >= 50on the other side of it.- Column
'A'starting with0seems to be a weird thing that needs special handling. It would be cleaner to assume that it begins with zero and starts incrementing at the firstTrue
In line with assign
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df.assign(A=c.cumsum(), B=(~c).cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 1 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 2 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 3 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 4 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 5 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 6 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 6 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 6 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 6 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 6 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 7 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 8 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 9 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 10 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 11 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 12 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 13 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 14 4
In place
a = df.f.values >= 50
b = df.C.values < 8
c = a | b
df[['A', 'B']] = np.column_stack([c, ~c]).cumsum(0)
df
Reduced
c = (df.f.values >= 50) | (df.C.values < 8)
df.assign(A=c.cumsum(), B=(~c).cumsum())
With special handling
a = df.f.values >= 50
b = df.C.values < 8
c0 = a | b
c1 = ~c0
c0[0] = False
c1[0] = False
df.assign(A=c0.cumsum(), B=c1.cumsum())
dtm f C A B
0 2018-03-01 00:00:00 +0000 50.135 9.0 0 0
1 2018-03-01 00:00:01 +0000 50.130 9.0 1 0
2 2018-03-01 00:00:02 +0000 50.120 9.0 2 0
3 2018-03-01 00:00:03 +0000 50.112 9.0 3 0
4 2018-03-01 00:00:04 +0000 50.102 9.0 4 0
5 2018-03-01 00:00:05 +0000 50.097 9.0 5 0
6 2018-03-01 00:00:06 +0000 11.095 9.0 5 1
7 2018-03-01 00:00:07 +0000 11.095 9.0 5 2
8 2018-03-01 00:00:08 +0000 11.092 9.0 5 3
9 2018-03-01 00:00:09 +0000 11.095 9.0 5 4
10 2018-03-01 00:00:10 +0000 11.097 5.0 6 4
11 2018-03-01 00:00:11 +0000 11.097 5.0 7 4
12 2018-03-01 00:00:12 +0000 11.097 5.0 8 4
13 2018-03-01 00:00:13 +0000 50.100 5.0 9 4
14 2018-03-01 00:00:14 +0000 50.102 5.0 10 4
15 2018-03-01 00:00:15 +0000 50.105 5.0 11 4
16 2018-03-01 00:00:16 +0000 50.102 5.0 12 4
17 2018-03-01 00:00:17 +0000 50.102 5.0 13 4
answered Nov 20 '18 at 15:30
piRSquaredpiRSquared
154k22146288
edited Nov 20 '18 at 15:56
answered Nov 20 '18 at 15:30
piRSquaredpiRSquared
154k22146288
answered Nov 20 '18 at 15:30
piRSquaredpiRSquared
154k22146288
answered Nov 20 '18 at 15:30
piRSquaredpiRSquared
154k22146288
154k22146288
add a comment |
add a comment |
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