Find consecutive ones in numpy array

问题

How can I find the amount of consecutive 1 (or any other value) in each row for of the following numpy array. I need a pure numpy solution.

counts
Out[304]: 
array([[0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0],
       [0, 0, 1, 0, 0, 1, 2, 0, 0, 1, 1, 1],
       [0, 0, 0, 4, 1, 0, 0, 0, 0, 1, 1, 0]])

desired solution first question (what is the maximum number of 1 in a row): amount: array([2,3,2])

second question (the index of where there are 2x a 1 in a row: index: array([3,9,9])

In this example I put 2x in a row. But it should be possible to change that to 5x in a row, this is important.

The second part of the question is, once it has found which ones have 5 or more consecutive 1 (or any other values), I will need the starting index of if. Again this should be done by row.

A similar question was answered with np.unique, but it only works for one row and not an array with multiple rows as the result would have different lengths: Get a list of all indices of repeated elements in a numpy array

回答1:

Here's a vectorized approach based on differentiation -

import numpy as np
import pandas  as pd

# Append zeros columns at either sides of counts
append1 = np.zeros((counts.shape[0],1),dtype=int)
counts_ext = np.column_stack((append1,counts,append1))

# Get start and stop indices with 1s as triggers
diffs = np.diff((counts_ext==1).astype(int),axis=1)
starts = np.argwhere(diffs == 1)
stops = np.argwhere(diffs == -1)

# Get intervals using differences between start and stop indices
start_stop = np.column_stack((starts[:,0], stops[:,1] - starts[:,1]))

# Get indices corresponding to max. interval lens and thus lens themselves
SS_df = pd.DataFrame(start_stop)
out = start_stop[SS_df.groupby([0],sort=False)[1].idxmax(),1]

Sample input, output -

Original sample case :

In [574]: counts
Out[574]: 
array([[0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0],
       [0, 0, 1, 0, 0, 1, 2, 0, 0, 1, 1, 1],
       [0, 0, 0, 4, 1, 0, 0, 0, 0, 1, 1, 0]])

In [575]: out
Out[575]: array([2, 3, 2], dtype=int64)

Modified case :

In [577]: counts
Out[577]: 
array([[0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0],
   [0, 0, 1, 0, 0, 1, 2, 0, 1, 1, 1, 1],
   [0, 0, 0, 4, 1, 1, 1, 1, 1, 0, 1, 0]])

In [578]: out
Out[578]: array([2, 4, 5], dtype=int64)

---

Here's a Pure NumPy version that is identical to the previous until we have start, stop. Here's the full implementation -

# Append zeros columns at either sides of counts
append1 = np.zeros((counts.shape[0],1),dtype=int)
counts_ext = np.column_stack((append1,counts,append1))

# Get start and stop indices with 1s as triggers
diffs = np.diff((counts_ext==1).astype(int),axis=1)
starts = np.argwhere(diffs == 1)
stops = np.argwhere(diffs == -1)

# Get intervals using differences between start and stop indices
intvs = stops[:,1] - starts[:,1]

# Store intervals as a 2D array for further vectorized ops to make.
c = np.bincount(starts[:,0])
mask = np.arange(c.max()) < c[:,None]
intvs2D = mask.astype(float)
intvs2D[mask] = intvs

# Get max along each row as final output
out = intvs2D.max(1)

回答2:

I think one problem that is very similar is to check if between the sorted rows the element wise difference is a certain amount. Here if there is a difference of 1 between 5 consecutive would be as follows. It can also be done for difference of 0 for two cards:

cardAmount=cards[0,:].size
has4=cards[:,np.arange(0,cardAmount-4)]-cards[:,np.arange(cardAmount-3,cardAmount)]
isStraight=np.any(has4 == 4, axis=1)

来源：https://stackoverflow.com/questions/35610834/find-consecutive-ones-in-numpy-array