How do I find: Is the first non-NaN value in each

问题:

For example:

      0     1
0  87.0   NaN
1   NaN  99.0
2   NaN   NaN
3   NaN   NaN
4   NaN  66.0
5   NaN   NaN
6   NaN  77.0
7   NaN   NaN
8   NaN   NaN
9  88.0   NaN

My expected output is: [False, True] since 87 is the first !NaN value but not the maximum in column 0. 99 however is the first !NaN value and is indeed the max in that column.

回答1:

Option a): Just do groupby with first

(May not be 100% reliable )

df.groupby([1]*len(df)).first()==df.max()
Out[89]: 
       0     1
1  False  True

Option b): bfill

Or using bfill(Fill any NaN value by the backward value in the column , then the first row after bfill is the first not NaN value )

df.bfill().iloc[0]==df.max()
Out[94]: 
0    False
1     True
dtype: bool

Option c): stack

df.stack().reset_index(level=1).drop_duplicates('level_1').set_index('level_1')[0]==df.max()
Out[102]: 
level_1
0    False
1     True
dtype: bool

Option d): idxmax with first_valid_index

df.idxmax()==df.apply(pd.Series.first_valid_index)
Out[105]: 
0    False
1     True
dtype: bool

Option e)(From Pir): idxmax with isna

df.notna().idxmax() == df.idxmax()     
Out[107]: 
0    False
1     True
dtype: bool

回答2:

Using pure numpy (I think this is very fast)

>>> np.isnan(df.values).argmin(axis=0) == df.fillna(-np.inf).values.argmax(axis=0)
array([False,  True])

The idea is to compare if the index of the first non-nan is also the index of the argmax.

Timings

df = pd.concat([df]*1000).reset_index(drop=True) # setup

%timeit np.isnan(df.values).argmin(axis=0) == df.fillna(-np.inf).values.argmax(axis=0)
207 µs ± 8.83 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit df.groupby([1]*len(df)).first()==df.max()
9.78 ms ± 339 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit df.bfill().iloc[0]==df.max()
824 µs ± 47.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit df.stack().reset_index(level=1).drop_duplicates('level_1').set_index('level_1')[0]==df.max()
3.55 ms ± 249 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit df.idxmax()==df.apply(pd.Series.first_valid_index)
1.5 ms ± 25 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit df.values[df.notnull().idxmax(), np.arange(df.shape[1])] == df.max(axis=0)
1.13 ms ± 14.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit df.values[(~np.isnan(df.values)).argmax(axis=0), np.arange(df.shape[1])] == df.max(axis=0).values
450 µs ± 20.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

回答3:

We can use numpy's nanmax here for an efficient solution:

a = df.values
np.nanmax(a, 0) == a[np.isnan(a).argmin(0), np.arange(a.shape[1])]

array([False,  True])

Timings (Whole lot of options presented here):

---

Functions

def chris(df):
    a = df.values
    return np.nanmax(a, 0) == a[np.isnan(a).argmin(0), np.arange(a.shape[1])]

def bradsolomon(df):
    df.values[df.notnull().idxmax(), np.arange(df.shape[1])] == df.max(axis=0).values

def wen1(df):
    return df.groupby([1]*len(df)).first()==df.max()

def wen2(df):
    return df.bfill().iloc[0]==df.max()

def wen3(df):
    return df.idxmax()==df.apply(pd.Series.first_valid_index)

def rafaelc(df):
    return np.isnan(df.values).argmin(axis=0) == df.fillna(-np.inf).values.argmax(axis=0)

def pir(df):
    return df.notna().idxmax() == df.idxmax()

Setup

res = pd.DataFrame(
       index=['chris', 'bradsolomon', 'wen1', 'wen2', 'wen3', 'rafaelc', 'pir'],
       columns=[10, 20, 30, 100, 500, 1000],
       dtype=float
)

for f in res.index:
    for c in res.columns:
        a = np.random.rand(c, c)
        a[a > 0.4] = np.nan
        df = pd.DataFrame(a)
        stmt = '{}(df)'.format(f)
        setp = 'from __main__ import df, {}'.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

回答4:

You can do something similar to Wens' answer with the underlying Numpy arrays:

>>> df.values[df.notnull().idxmax(), np.arange(df.shape[1])] == df.max(axis=0).values
array([False,  True])

df.max(axis=0) gives the column-wise max.

The left hand side indexes df.values, which is a 2d array, to make it a 1d array and compare it element-wise to the maxes per column.

If you exclude .values from the right-hand side, the result will just be a Pandas Series:

>>> df.values[df.notnull().idxmax(), np.arange(df.shape[1])] == df.max(axis=0)
0    False
1     True
dtype: bool

回答5:

After posting the question I came up with this:

def nice_method_name_here(sr):
    return sr[sr > 0][0] == np.max(sr)

print(df.apply(nice_method_name_here))

which seems to work, but not sure yet!