Great question—this is a super common pain point when working with missing data in NumPy, especially for domain-specific tasks like your ice volume calculations where a spurious 0 can throw off your analysis entirely. Let’s break down your two questions:

1. Why doesn’t np.nansum() have a keyword argument to return nan for all-NaN arrays? #

NumPy’s nan-aware functions follow mathematical conventions first:

• For nanmean, when all elements are NaN, there are no valid values to compute a mean from—so it returns nan (since dividing by zero is undefined).
• For nansum, the sum of an empty set (which is what you’re left with when all elements are NaN) is mathematically defined as 0. This aligns with standard math and programming norms (like how an empty list’s sum is 0 in vanilla Python).

That said, I totally get why this is frustrating for your use case—0 is a meaningful value (actual zero ice coverage) whereas all NaNs means missing data, and you need to distinguish those two scenarios clearly.

Unfortunately, NumPy doesn’t have a built-in keyword argument to override this behavior for nansum right now. The good news is you can easily wrap the function to add this logic yourself (we’ll cover that in a bit, alongside checking for all-NaN arrays).

2. Is there a function to check if an entire matrix is all NaN? #

Absolutely! You can combine np.isnan() (which returns a boolean array marking NaN positions) with np.all() (which verifies if every element in the array is True):

import numpy as np

# For a 1D array
X = np.array([np.nan, np.nan, np.nan])
print(np.all(np.isnan(X)))  # Output: True

# For your 180x360 ice data array
ice_data = np.full((180, 360), np.nan)
print(np.all(np.isnan(ice_data)))  # Output: True

Putting it all together for your ice data workflow #

To get nan instead of 0 when your array is entirely missing data, create a simple helper function:

def safe_nansum(arr):
    if np.all(np.isnan(arr)):
        return np.nan
    return np.nansum(arr)

# Test with all NaNs (missing data)
all_nan_ice = np.full((180, 360), np.nan)
print(safe_nansum(all_nan_ice))  # Output: nan

# Test with actual ice coverage (mix of NaNs and valid values)
real_ice = np.full((180, 360), np.nan)
real_ice[50:100, 100:200] = 1  # Simulate 50x100 km of ice
print(safe_nansum(real_ice))  # Output: 5000.0 (your valid sum)

This way, you’ll only get 0 when there’s actual zero ice coverage (valid values that sum to 0) and nan when the entire dataset is missing—exactly what you need to avoid misleading results.

内容的提问来源于stack exchange，提问作者J W