Remove leading NaN in pandas

Question

How can I remove leading NaN's in pandas?

pd.Series([np.nan, np.nan, np.nan, 1, 2, np.nan, 3])

I want to remove only the first 3 NaN's from above, so the result should be:

pd.Series([1, 2, np.nan, 3])

Answer 1

Here is another method using pandas methods only:

In [103]:
s = pd.Series([np.nan, np.nan, np.nan, 1, 2, np.nan, 3])
first_valid = s[s.notnull()].index[0]
s.iloc[first_valid:]

Out[103]:
3     1
4     2
5   NaN
6     3
dtype: float64

So we filter the series using notnull to get the first valid index. Then use iloc to slice the series

EDIT

As @ajcr has pointed out it is better to use the built-in method first_valid_index as this does not return a temp series which I'm using to mask in the above answer, additionally using loc uses the index label rather than iloc which uses ordinal position which works for the general case where the index is not an int64Index:

In [104]:
s = pd.Series([np.nan, np.nan, np.nan, 1, 2, np.nan, 3])
s.loc[s.first_valid_index():]

Out[104]:
3     1
4     2
5   NaN
6     3
dtype: float64

Answer 2

Find first non-nan index

To find the index of the first non-nan item

s = pd.Series([np.nan, np.nan, np.nan, 1, 2, np.nan, 3])

nans = s.apply(np.isnan)

first_non_nan = nans[nans == False].index[0] # get the first one

Output

s[first_non_nan:]
Out[44]:
3     1
4     2
5   NaN
6     3
dtype: float64

Answer 3

Two more approaches could be suggested here, assuming A as the input series.

Approach #1: With slicing -

A[np.where(~np.isnan(A))[0][0]:] 

Approach #2: With masking -

A[np.maximum.accumulate(~np.isnan(A))]

Sample run -

In [219]: A = pd.Series([np.nan, np.nan, np.nan, 1, 2, np.nan, 3])

In [220]: A
Out[220]: 
0   NaN
1   NaN
2   NaN
3     1
4     2
5   NaN
6     3
dtype: float64

In [221]: A[np.where(~np.isnan(A))[0][0]:]       # Approach 1
Out[221]: 
3     1
4     2
5   NaN
6     3
dtype: float64

In [222]: A[np.maximum.accumulate(~np.isnan(A))]  # Approach 2
Out[222]: 
3     1
4     2
5   NaN
6     3
dtype: float64