The original dataframe is a table like this:
S1_r1_ctrl/ S1_r2_ctrl/ S1_r3_ctrl/
sp|P38646|GRP75_HUMAN 2.960000e-06 5.680000e-06 0.000000e+00
sp|O75694-2|NU155_HUMAN 2.710000e-07 0.000000e+00 2.180000e-07
sp|Q05397-2|FAK1_HUMAN 0.000000e+00 2.380000e-07 7.330000e-06
sp|O60671-2|RAD1_HUMAN NaN NaN NaN
I am looking for the smallest value in each column of a dataframe greater than zero. I was trying to use this example to answer my question. My code looks like:
df.ne(0).idxmin().to_frame('pos').assign(value=lambda d: df.lookup(d.pos, d.index))
but still I get only zeros and my result looks like this:
pos value
S1_r1_ctrl/ sp|Q05397-2|FAK1_HUMAN 0.0
S1_r2_ctrl/ sp|O75694-2|NU155_HUMAN 0.0
S1_r3_ctrl/ sp|P38646|GRP75_HUMAN 0.0
instead of this:
pos value
S1_r1_ctrl/ sp|O75694-2|NU155_HUMAN 2.710000e-07
S1_r2_ctrl/ sp|Q05397-2|FAK1_HUMAN 2.380000e-07
S1_r3_ctrl/ sp|O75694-2|NU155_HUMAN 2.180000e-07
I guess there might be a problem in data types but I'm not sure. I assumed ne(0)
would ignore zeros but it doesn't so I am confused why. And perhaps there's a more intelligent way to find what I need.
Setup
df = pd.DataFrame([[0, 0, 0],
[0, 10, 0],
[4, 0, 0],
[1, 2, 3]],
columns=['first', 'second', 'third'])
Using a mask with min(0)
:
df[df.gt(0)].min(0)
first 1.0
second 2.0
third 3.0
dtype: float64
As @DSM pointed out, this can also be written:
df.where(df.gt(0)).min(0)
Performance
def chris():
df1[df1.gt(0)].min(0)
def chris2():
df1.where(df1.gt(0)).min(0)
def wen():
a=df1.values.T
a = np.ma.masked_equal(a, 0.0, copy=False)
a.min(1)
def haleemur():
df1.replace(0, np.nan).min()
Setup
from timeit import timeit
import matplotlib.pyplot as plt
res = pd.DataFrame(
index=['chris', 'chris2', 'wen', 'haleemur'],
columns=[10, 50, 100, 500, 1000, 5000, 10000, 50000, 100000],
dtype=float
)
for f in res.index:
for c in res.columns:
df1 = df.copy()
df1 = pd.concat([df1]*c)
stmt = '{}()'.format(f)
setp = 'from __main__ import df1, {}'.format(f)
res.at[f, c] = timeit(stmt, setp, number=50)
ax = res.div(res.min()).T.plot(loglog=True)
ax.set_xlabel("N");
ax.set_ylabel("time (relative)");
plt.show()
Results
Maybe numpy
is good alternative
a=df.values.T
a = np.ma.masked_equal(a, 0.0, copy=False)
a.min(1)
Out[755]:
masked_array(data=[1, 2, 3],
mask=[False, False, False],
fill_value=999999,
dtype=int64)
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