Return dataframe subset based on a list of boolean values

Question

I\'m trying to slice a dataframe based on list of values, how would I go about this?

Say I have an expression or a list l = [0,1,0,0,1,1,0,0,0,1]

Ho

Answer 1

Convert the list to a boolean array and then use boolean indexing:

df = pd.DataFrame(np.random.randint(10, size=(10, 3)))

df[np.array(lst).astype(bool)]
Out: 
   0  1  2
1  8  6  3
4  2  7  3
5  7  2  3
9  1  3  4

Answer 2

yet another "creative" approach:

In [181]: a = np.array(lst)

In [182]: df.query("index * @a > 0")
Out[182]:
   0  1  2
1  1  5  5
4  0  2  0
5  4  9  9
9  2  2  5

or much better variant from @ayhan:

In [183]: df.query("@a != 0")
Out[183]:
   0  1  2
1  1  5  5
4  0  2  0
5  4  9  9
9  2  2  5

PS i've also borrowed @Ayhan's setup

Answer 3

Or maybe find the position of 1 in your list and slice from the Dataframe

df.loc[[i for i,x in enumerate(lst) if x == 1],:]

Answer 4

Selecting using a list of Booleans is something itertools.compress does well.

Given

>>> df = pd.DataFrame(np.random.randint(10, size=(10, 2)))
>>> selectors = [0, 1, 0, 0, 1, 1, 0, 0, 0, 1]

Code

>>> selected_idxs = list(itertools.compress(df.index, selectors))   # [1, 4, 5, 9]
>>> df.iloc[selected_idxs, :]
   0  1
1  1  9
4  3  4
5  4  1
9  8  9

Answer 5

Setup
Borrowed @ayhan's setup

df = pd.DataFrame(np.random.randint(10, size=(10, 3)))

Without numpy
not the fastest, but it holds its own and is definitely the shortest.

df[list(map(bool, lst))]

   0  1  2
1  3  5  6
4  6  3  2
5  5  7  6
9  0  0  1

---

Timing

results.div(results.min(1), 0).round(2).pipe(lambda d: d.assign(Best=d.idxmin(1)))

         ayh   wvo   pir   mxu   wen Best
N                                        
1       1.53  1.00  1.02  4.95  2.61  wvo
3       1.06  1.00  1.04  5.46  2.84  wvo
10      1.00  1.00  1.00  4.30  2.73  ayh
30      1.00  1.05  1.24  4.06  3.76  ayh
100     1.16  1.00  1.19  3.90  3.53  wvo
300     1.29  1.00  1.32  2.50  2.38  wvo
1000    1.54  1.00  2.19  2.24  3.85  wvo
3000    1.39  1.00  2.17  1.81  4.55  wvo
10000   1.22  1.00  2.21  1.35  4.36  wvo
30000   1.19  1.00  2.26  1.39  5.36  wvo
100000  1.19  1.00  2.19  1.31  4.82  wvo

---

fig, (a1, a2) = plt.subplots(2, 1, figsize=(6, 6))
results.plot(loglog=True, lw=3, ax=a1)
results.div(results.min(1), 0).round(2).plot.bar(logy=True, ax=a2)
fig.tight_layout()

---

Testing Code

ayh = lambda d, l: d[np.array(l).astype(bool)]
wvo = lambda d, l: d[np.array(l, dtype=bool)]
pir = lambda d, l: d[list(map(bool, l))]
wen = lambda d, l: d.loc[[i for i, x in enumerate(l) if x == 1], :]

def mxu(d, l):
    a = np.array(l)
    return d.query('@a != 0')

results = pd.DataFrame(
    index=pd.Index([1, 3, 10, 30, 100, 300,
                    1000, 3000, 10000, 30000, 100000], name='N'),
    columns='ayh wvo pir mxu wen'.split(),
    dtype=float
)

for i in results.index:
    d = pd.concat([df] * i, ignore_index=True)
    l = lst * i
    for j in results.columns:
        stmt = '{}(d, l)'.format(j)
        setp = 'from __main__ import d, l, {}'.format(j)
        results.set_value(i, j, timeit(stmt, setp, number=10))

Answer 6

You can use masking here:

df[np.array([0,1,0,0,1,1,0,0,0,1],dtype=bool)]

So we construct a boolean array with true and false. Every place where the array is True is a row we select.

Mind that we do not filter inplace. In order to retrieve the result, you have to assign the result to an (optionally different) variable:

df2 = df[np.array([0,1,0,0,1,1,0,0,0,1],dtype=bool)]