Pandas: Count the first consecutive True values

Question:

Find the count of the first consecutive Trues
Consider a

a = np.array([True, True, True, False, True, False, True, True, True, True])  

Answer 1
numpy: Use np.logical_and.accumulate on the negation of a and take the negation of that to make a mask that eliminates the first series of Falses if they should exist. Then append a False at the end to ensure we have a non True min. Finally, use np.argmin to locate the first minimum value. If it's found a position 3, that will indicate 3 True values before it.

np.argmin(np.append(a[~np.logical_and.accumulate(~a)], False))

3

---

Answer 2
numba.njit

I'd like to use numba so I can loop and make sure I get to short circuit when we want/need to. This is a problem that is sure to be answered early in the array. There isn't need to evaluate things along the entire array for no reason.

from numba import njit

@njit
def first_true(a):
    true_started = False
    c = 0
    for i, j in enumerate(a):
        if true_started and not j:
            return c
        else:
            c += j
            true_started = true_started or j
    return c

first_true(a)

3

---

Answer 3
numpy smarter use of argmin and argmax. I surround a with False then use argmax to find the first True then from that point on, use argmin to find the first False after that.
Note: @Divakar made an improvement on this answer that eliminates the use of np.concatenate and uses if/then/else instead. That cut this already very fast solution by a factor of 3!

def first_true2(a):
    a = np.concatenate([[False], a, [False]])
    return np.argmin(a[np.argmax(a):])

first_true2(a)

3

---

How fast are these answers?
See @Divakar's Answer for source code of other functions being timed

%timeit first_true(a)
%timeit np.argmin(np.append(a[~np.logical_and.accumulate(~a)], False))
%timeit np.diff(np.flatnonzero(np.diff(np.r_[0,a,0])))[0]
%timeit first_True_island_len(a)
%timeit first_true2(a)
%timeit first_True_island_len_IFELSE(a)


a = np.array([True, True, True, False, True, False, True, True, True, True])    
1000000 loops, best of 3: 353 ns per loop
100000 loops, best of 3: 8.32 µs per loop
10000 loops, best of 3: 27.4 µs per loop
100000 loops, best of 3: 5.48 µs per loop
100000 loops, best of 3: 5.38 µs per loop
1000000 loops, best of 3: 1.35 µs per loop

a = np.array([False] * 100000 + [True] * 10000)
10000 loops, best of 3: 112 µs per loop
10000 loops, best of 3: 127 µs per loop
1000 loops, best of 3: 513 µs per loop
10000 loops, best of 3: 110 µs per loop
100000 loops, best of 3: 13.9 µs per loop
100000 loops, best of 3: 4.55 µs per loop

a = np.array([False] * 100000 + [True])
10000 loops, best of 3: 102 µs per loop
10000 loops, best of 3: 115 µs per loop
1000 loops, best of 3: 472 µs per loop
10000 loops, best of 3: 108 µs per loop
100000 loops, best of 3: 14 µs per loop
100000 loops, best of 3: 4.45 µs per loop

Using NumPy funcs, one solution would be -

np.diff(np.flatnonzero(np.diff(np.r_[0,s,0])))[0]

Sample run -

In [16]: s
Out[16]: 
0     True
1     True
2     True
3    False
4     True
5    False
6     True
7     True
8     True
9     True
dtype: bool

In [17]: np.diff(np.flatnonzero(np.diff(np.r_[0,s,0])))[0]
Out[17]: 3

For performance, we need to use np.concatenate in place np.r_ and then slicing to replace the last differentiation -

def first_True_island_len(a): # a is NumPy array
    v = np.concatenate(([False],a,[False]))
    idx = np.flatnonzero(v[1:] != v[:-1])
    if len(idx)>0:
        return idx[1] - idx[0]
    else:
        return 0

Inspired by @piRSquared's argmax and argmin trickery, here's one more with a bunch of IF-ELSE's -

def first_True_island_len_IFELSE(a): # a is NumPy array
    maxidx = a.argmax()
    pos = a[maxidx:].argmin()
    if a[maxidx]:
        if pos==0:
            return a.size - maxidx
        else:
            return pos
    else:
        return 0

Try this way will find the first consecutive occurrences for True or False, and only for True:

import pandas as pd
df = pd.DataFrame([True, True, True, False, True, False, True, True, True, True],columns=["Boolean"])
df['consecutive'] = df.Boolean.groupby((df.Boolean != df.Boolean.shift()).cumsum()).transform('size')
count_true_false = df['consecutive'][df['consecutive']>1].iloc[0] # count first consecutive occurrences for True or False 
count_true = df['consecutive'][(df.Boolean == True) & (df.consecutive>1)].iloc[0] # count first consecutive occurrences for True 
print count_true_false
print count_true

Output:

3
3