How to count occurence of true positives using pandas or numpy?

I have two columns, Prediction and Ground Truth.
I want to get a count of true positives as a series using either numpy or pandas.

For example, my data is:

Prediction GroundTruth

True       True

True       False

True       True

False      True

False      False

True       True

I want a list that should have the following output:

tp_list = [1,1,2,2,2,3]

Is there a one-liner way to do this in numpy or pandas?

Currently, this is my solution:

tp = 0

for p, g in zip(data.Prediction, data.GroundTruth):

  if p and g: # TP case

    tp = tp + 1

  tp_list.append(tp)

edited Nov 21 at 2:36

RafaelC

25.9k82649

asked Nov 21 at 2:18

Raashid

325

add a comment |

I have two columns, Prediction and Ground Truth.
I want to get a count of true positives as a series using either numpy or pandas.

For example, my data is:

Prediction GroundTruth

True       True

True       False

True       True

False      True

False      False

True       True

I want a list that should have the following output:

tp_list = [1,1,2,2,2,3]

Is there a one-liner way to do this in numpy or pandas?

Currently, this is my solution:

tp = 0

for p, g in zip(data.Prediction, data.GroundTruth):

  if p and g: # TP case

    tp = tp + 1

  tp_list.append(tp)

edited Nov 21 at 2:36

RafaelC

25.9k82649

asked Nov 21 at 2:18

Raashid

325

add a comment |

I have two columns, Prediction and Ground Truth.
I want to get a count of true positives as a series using either numpy or pandas.

For example, my data is:

Prediction GroundTruth

True       True

True       False

True       True

False      True

False      False

True       True

I want a list that should have the following output:

tp_list = [1,1,2,2,2,3]

Is there a one-liner way to do this in numpy or pandas?

Currently, this is my solution:

tp = 0

for p, g in zip(data.Prediction, data.GroundTruth):

  if p and g: # TP case

    tp = tp + 1

  tp_list.append(tp)

edited Nov 21 at 2:36

RafaelC

25.9k82649

asked Nov 21 at 2:18

Raashid

325

I have two columns, Prediction and Ground Truth.
I want to get a count of true positives as a series using either numpy or pandas.

For example, my data is:

Prediction GroundTruth

True       True

True       False

True       True

False      True

False      False

True       True

I want a list that should have the following output:

tp_list = [1,1,2,2,2,3]

Is there a one-liner way to do this in numpy or pandas?

Currently, this is my solution:

tp = 0

for p, g in zip(data.Prediction, data.GroundTruth):

  if p and g: # TP case

    tp = tp + 1

  tp_list.append(tp)

python pandas numpy

edited Nov 21 at 2:36

RafaelC

25.9k82649

asked Nov 21 at 2:18

Raashid

325

edited Nov 21 at 2:36

RafaelC

25.9k82649

asked Nov 21 at 2:18

Raashid

325

edited Nov 21 at 2:36

RafaelC

25.9k82649

edited Nov 21 at 2:36

RafaelC

25.9k82649

edited Nov 21 at 2:36

RafaelC

25.9k82649

asked Nov 21 at 2:18

Raashid

325

asked Nov 21 at 2:18

Raashid

325

asked Nov 21 at 2:18

Raashid

325

add a comment |

3 Answers
3

active

oldest

votes

To get a running count (i.e., cumulative sum) of true positives, i.e., Prediction == True if and only if GroundTruth == True, the solution is a modification of @RafaelC's answer:

(df['Prediction'] & df['GroundTruth']).cumsum()

0    1

1    1

2    2

3    2

4    2

5    3



(df['Prediction'] & df['GroundTruth']).cumsum().tolist()

[1, 1, 2, 2, 2, 3]

answered Nov 21 at 2:29

Peter Leimbigler

3,7041415

add a comment |

If you want to know how many True you predicted that are actually True, use

(df['Prediction'] & df['GroundTruth']).cumsum()



0    1

1    1

2    2

3    2

4    2

5    3

dtype: int64

(thanks @Peter Leimbigiler for chiming in)

If you want to know how many you have predicted correctly just compare and use cumsum

(df['Prediction'] == df['GroundTruth']).cumsum()

which outputs

0    1

1    1

2    2

3    2

4    3

5    4

dtype: int64

Can always get a list by using .tolist()

(df4['Prediction'] == df4['GroundTruth']).cumsum().tolist()



[1, 1, 2, 2, 3, 4]

edited Nov 21 at 2:29

answered Nov 21 at 2:20

RafaelC

25.9k82649

df['Prediction'] == df['GroundTruth'] gives true positives plus true negatives, basically the accuracy score before dividing by the number of predictions. The running count of true positives is given by df['Prediction'] & df['GroundTruth'].
– Peter Leimbigler
Nov 21 at 2:22

if the ground truth is False and the prediction is False, what do you call this outcome? The predictor correctly classified a negative instance. It's a true negative.
– Peter Leimbigler
Nov 21 at 2:25

2

@RafaelC, the distinction between "correct prediction" and "correct prediction of a negative outcome" is important :) en.wikipedia.org/wiki/Confusion_matrix
– Peter Leimbigler
Nov 21 at 2:30

1

Thanks for chiming in guys :) Got the point
– RafaelC
Nov 21 at 2:31

1

Thanks a ton guys! you are too fast. I posted this question right before boarding a flight and I had an answer after boarding!
– Raashid
Nov 21 at 13:42

add a comment |

Maybe you can using all

df.all(1).cumsum().tolist()

Out[156]: [1, 1, 2, 2, 2, 3]

numpy solution

np.cumsum(np.all(df.values,1))

Out[159]: array([1, 1, 2, 2, 2, 3], dtype=int32)

answered Nov 21 at 2:48

W-B

99.8k73163

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

To get a running count (i.e., cumulative sum) of true positives, i.e., Prediction == True if and only if GroundTruth == True, the solution is a modification of @RafaelC's answer:

(df['Prediction'] & df['GroundTruth']).cumsum()

0    1

1    1

2    2

3    2

4    2

5    3



(df['Prediction'] & df['GroundTruth']).cumsum().tolist()

[1, 1, 2, 2, 2, 3]

answered Nov 21 at 2:29

Peter Leimbigler

3,7041415

add a comment |

To get a running count (i.e., cumulative sum) of true positives, i.e., Prediction == True if and only if GroundTruth == True, the solution is a modification of @RafaelC's answer:

(df['Prediction'] & df['GroundTruth']).cumsum()

0    1

1    1

2    2

3    2

4    2

5    3



(df['Prediction'] & df['GroundTruth']).cumsum().tolist()

[1, 1, 2, 2, 2, 3]

answered Nov 21 at 2:29

Peter Leimbigler

3,7041415

add a comment |

To get a running count (i.e., cumulative sum) of true positives, i.e., Prediction == True if and only if GroundTruth == True, the solution is a modification of @RafaelC's answer:

(df['Prediction'] & df['GroundTruth']).cumsum()

0    1

1    1

2    2

3    2

4    2

5    3



(df['Prediction'] & df['GroundTruth']).cumsum().tolist()

[1, 1, 2, 2, 2, 3]

answered Nov 21 at 2:29

Peter Leimbigler

3,7041415

To get a running count (i.e., cumulative sum) of true positives, i.e., Prediction == True if and only if GroundTruth == True, the solution is a modification of @RafaelC's answer:

(df['Prediction'] & df['GroundTruth']).cumsum()

0    1

1    1

2    2

3    2

4    2

5    3



(df['Prediction'] & df['GroundTruth']).cumsum().tolist()

[1, 1, 2, 2, 2, 3]

answered Nov 21 at 2:29

Peter Leimbigler

3,7041415

answered Nov 21 at 2:29

Peter Leimbigler

3,7041415

answered Nov 21 at 2:29

Peter Leimbigler

3,7041415

answered Nov 21 at 2:29

Peter Leimbigler

3,7041415

add a comment |

If you want to know how many True you predicted that are actually True, use

(df['Prediction'] & df['GroundTruth']).cumsum()



0    1

1    1

2    2

3    2

4    2

5    3

dtype: int64

(thanks @Peter Leimbigiler for chiming in)

If you want to know how many you have predicted correctly just compare and use cumsum

(df['Prediction'] == df['GroundTruth']).cumsum()

which outputs

0    1

1    1

2    2

3    2

4    3

5    4

dtype: int64

Can always get a list by using .tolist()

(df4['Prediction'] == df4['GroundTruth']).cumsum().tolist()



[1, 1, 2, 2, 3, 4]

edited Nov 21 at 2:29

answered Nov 21 at 2:20

RafaelC

25.9k82649

df['Prediction'] == df['GroundTruth'] gives true positives plus true negatives, basically the accuracy score before dividing by the number of predictions. The running count of true positives is given by df['Prediction'] & df['GroundTruth'].
– Peter Leimbigler
Nov 21 at 2:22

if the ground truth is False and the prediction is False, what do you call this outcome? The predictor correctly classified a negative instance. It's a true negative.
– Peter Leimbigler
Nov 21 at 2:25

2

@RafaelC, the distinction between "correct prediction" and "correct prediction of a negative outcome" is important :) en.wikipedia.org/wiki/Confusion_matrix
– Peter Leimbigler
Nov 21 at 2:30

1

Thanks for chiming in guys :) Got the point
– RafaelC
Nov 21 at 2:31

1

Thanks a ton guys! you are too fast. I posted this question right before boarding a flight and I had an answer after boarding!
– Raashid
Nov 21 at 13:42

add a comment |

If you want to know how many True you predicted that are actually True, use

(df['Prediction'] & df['GroundTruth']).cumsum()



0    1

1    1

2    2

3    2

4    2

5    3

dtype: int64

(thanks @Peter Leimbigiler for chiming in)

If you want to know how many you have predicted correctly just compare and use cumsum

(df['Prediction'] == df['GroundTruth']).cumsum()

which outputs

0    1

1    1

2    2

3    2

4    3

5    4

dtype: int64

Can always get a list by using .tolist()

(df4['Prediction'] == df4['GroundTruth']).cumsum().tolist()



[1, 1, 2, 2, 3, 4]

edited Nov 21 at 2:29

answered Nov 21 at 2:20

RafaelC

25.9k82649

df['Prediction'] == df['GroundTruth'] gives true positives plus true negatives, basically the accuracy score before dividing by the number of predictions. The running count of true positives is given by df['Prediction'] & df['GroundTruth'].
– Peter Leimbigler
Nov 21 at 2:22

if the ground truth is False and the prediction is False, what do you call this outcome? The predictor correctly classified a negative instance. It's a true negative.
– Peter Leimbigler
Nov 21 at 2:25

2

@RafaelC, the distinction between "correct prediction" and "correct prediction of a negative outcome" is important :) en.wikipedia.org/wiki/Confusion_matrix
– Peter Leimbigler
Nov 21 at 2:30

1

Thanks for chiming in guys :) Got the point
– RafaelC
Nov 21 at 2:31

1

Thanks a ton guys! you are too fast. I posted this question right before boarding a flight and I had an answer after boarding!
– Raashid
Nov 21 at 13:42

add a comment |

If you want to know how many True you predicted that are actually True, use

(df['Prediction'] & df['GroundTruth']).cumsum()



0    1

1    1

2    2

3    2

4    2

5    3

dtype: int64

(thanks @Peter Leimbigiler for chiming in)

If you want to know how many you have predicted correctly just compare and use cumsum

(df['Prediction'] == df['GroundTruth']).cumsum()

which outputs

0    1

1    1

2    2

3    2

4    3

5    4

dtype: int64

Can always get a list by using .tolist()

(df4['Prediction'] == df4['GroundTruth']).cumsum().tolist()



[1, 1, 2, 2, 3, 4]

edited Nov 21 at 2:29

answered Nov 21 at 2:20

RafaelC

25.9k82649

If you want to know how many True you predicted that are actually True, use

(df['Prediction'] & df['GroundTruth']).cumsum()



0    1

1    1

2    2

3    2

4    2

5    3

dtype: int64

(thanks @Peter Leimbigiler for chiming in)

If you want to know how many you have predicted correctly just compare and use cumsum

(df['Prediction'] == df['GroundTruth']).cumsum()

which outputs

0    1

1    1

2    2

3    2

4    3

5    4

dtype: int64

Can always get a list by using .tolist()

(df4['Prediction'] == df4['GroundTruth']).cumsum().tolist()



[1, 1, 2, 2, 3, 4]

edited Nov 21 at 2:29

answered Nov 21 at 2:20

RafaelC

25.9k82649

edited Nov 21 at 2:29

answered Nov 21 at 2:20

RafaelC

25.9k82649

answered Nov 21 at 2:20

RafaelC

25.9k82649

answered Nov 21 at 2:20

RafaelC

25.9k82649

df['Prediction'] == df['GroundTruth'] gives true positives plus true negatives, basically the accuracy score before dividing by the number of predictions. The running count of true positives is given by df['Prediction'] & df['GroundTruth'].
– Peter Leimbigler
Nov 21 at 2:22

if the ground truth is False and the prediction is False, what do you call this outcome? The predictor correctly classified a negative instance. It's a true negative.
– Peter Leimbigler
Nov 21 at 2:25

2

@RafaelC, the distinction between "correct prediction" and "correct prediction of a negative outcome" is important :) en.wikipedia.org/wiki/Confusion_matrix
– Peter Leimbigler
Nov 21 at 2:30

1

Thanks for chiming in guys :) Got the point
– RafaelC
Nov 21 at 2:31

1

Thanks a ton guys! you are too fast. I posted this question right before boarding a flight and I had an answer after boarding!
– Raashid
Nov 21 at 13:42

add a comment |

df['Prediction'] == df['GroundTruth'] gives true positives plus true negatives, basically the accuracy score before dividing by the number of predictions. The running count of true positives is given by df['Prediction'] & df['GroundTruth'].
– Peter Leimbigler
Nov 21 at 2:22

if the ground truth is False and the prediction is False, what do you call this outcome? The predictor correctly classified a negative instance. It's a true negative.
– Peter Leimbigler
Nov 21 at 2:25

2

@RafaelC, the distinction between "correct prediction" and "correct prediction of a negative outcome" is important :) en.wikipedia.org/wiki/Confusion_matrix
– Peter Leimbigler
Nov 21 at 2:30

1

Thanks for chiming in guys :) Got the point
– RafaelC
Nov 21 at 2:31

1

Thanks a ton guys! you are too fast. I posted this question right before boarding a flight and I had an answer after boarding!
– Raashid
Nov 21 at 13:42

df['Prediction'] == df['GroundTruth'] gives true positives plus true negatives, basically the accuracy score before dividing by the number of predictions. The running count of true positives is given by df['Prediction'] & df['GroundTruth'].
– Peter Leimbigler
Nov 21 at 2:22

if the ground truth is False and the prediction is False, what do you call this outcome? The predictor correctly classified a negative instance. It's a true negative.
– Peter Leimbigler
Nov 21 at 2:25

@RafaelC, the distinction between "correct prediction" and "correct prediction of a negative outcome" is important :) en.wikipedia.org/wiki/Confusion_matrix
– Peter Leimbigler
Nov 21 at 2:30

Thanks for chiming in guys :) Got the point
– RafaelC
Nov 21 at 2:31

Thanks a ton guys! you are too fast. I posted this question right before boarding a flight and I had an answer after boarding!
– Raashid
Nov 21 at 13:42

add a comment |

Maybe you can using all

df.all(1).cumsum().tolist()

Out[156]: [1, 1, 2, 2, 2, 3]

numpy solution

np.cumsum(np.all(df.values,1))

Out[159]: array([1, 1, 2, 2, 2, 3], dtype=int32)

answered Nov 21 at 2:48

W-B

99.8k73163

add a comment |

Maybe you can using all

df.all(1).cumsum().tolist()

Out[156]: [1, 1, 2, 2, 2, 3]

numpy solution

np.cumsum(np.all(df.values,1))

Out[159]: array([1, 1, 2, 2, 2, 3], dtype=int32)

answered Nov 21 at 2:48

W-B

99.8k73163

add a comment |

Maybe you can using all

df.all(1).cumsum().tolist()

Out[156]: [1, 1, 2, 2, 2, 3]

numpy solution

np.cumsum(np.all(df.values,1))

Out[159]: array([1, 1, 2, 2, 2, 3], dtype=int32)

answered Nov 21 at 2:48

W-B

99.8k73163

Maybe you can using all

df.all(1).cumsum().tolist()

Out[156]: [1, 1, 2, 2, 2, 3]

numpy solution

np.cumsum(np.all(df.values,1))

Out[159]: array([1, 1, 2, 2, 2, 3], dtype=int32)

answered Nov 21 at 2:48

W-B

99.8k73163

answered Nov 21 at 2:48

W-B

99.8k73163

answered Nov 21 at 2:48

W-B

99.8k73163

answered Nov 21 at 2:48

W-B

99.8k73163

add a comment |

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