Pandas count the number of elements in each column less than x

Question

I have a DataFrame that looks below. I try to count the number of elements less than 2.0 in each column, then I will visualize the result on the graph. I did this with lists and loops, but I'm wondering if there is a “Pandas” way to do this quickly. Thanks!

x = [] for i in range(6): x.append(df[df.ix[:,i]<2.0].count()[i]) 

then I can get the bar chart using the list x .

  ABCDEF 0 2.142 1.929 1.674 1.547 3.395 2.382 1 2.077 1.871 1.614 1.491 3.110 2.288 2 2.098 1.889 1.610 1.487 3.020 2.262 3 1.990 1.760 1.479 1.366 2.496 2.128 4 1.935 1.765 1.656 1.530 2.786 2.433 

Answer 1

 In [96]: df = pd.DataFrame({'a':randn(10), 'b':randn(10), 'c':randn(10)}) df Out[96]: abc 0 -0.849903 0.944912 1.285790 1 -1.038706 1.445381 0.251002 2 0.683135 -0.539052 -0.622439 3 -1.224699 -0.358541 1.361618 4 -0.087021 0.041524 0.151286 5 -0.114031 -0.201018 -0.030050 6 0.001891 1.601687 -0.040442 7 0.024954 -1.839793 0.917328 8 -1.480281 0.079342 -0.405370 9 0.167295 -1.723555 -0.033937 [10 rows x 3 columns] In [97]: df[df > 1.0].count() Out[97]: a 0 b 2 c 2 dtype: int64 

So in your case:

 df[df < 2.0 ].count() 

must work

EDIT

some timings

 In [3]: %timeit df[df < 1.0 ].count() %timeit (df < 1.0).sum() %timeit (df < 1.0).apply(np.count_nonzero) 1000 loops, best of 3: 1.47 ms per loop 1000 loops, best of 3: 560 us per loop 1000 loops, best of 3: 529 us per loop 

So @DSM's suggestions are true and much faster than my suggestion