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[英]Replace nan's for a list according to another list with different sizes python
[英]How to replace elements in a or more list to nan by list nan's position in python3?
a = [1,2,3,np.nan,4,6,8,np.nan,5,3,7]
b = [np.nan,2,3,5,7,np.nan,6,2,6,9]
for i in np.arange(0,12,1):
if a[i] == np.nan or b[i]==np.nan:
a[i] = np.nan
b[i] =np.nan
我是 python 的新手,我想要最终结果:
a = [np.nan, 2,3,np.nan, 4, 6, np.nan, np.nan,5,3,7]
b = [np.nan,2,3,np.nan,7,np.nan,np.nan,2,6,9]
但它没有用。 非常感谢您的任何建议
您正在使用 numpy,那么为什么不使用 numpy arrays?
a, b = map(np.array, [a, b])
l = min(a.size, b.size)
# Get a mask of NaN cells
m = np.isnan(a[:l]) | np.isnan(b[:l])
# Set to NaN based on the mask
a[:l][m] = np.nan
b[:l][m] = np.nan
a
# array([nan, 2., 3., nan, 4., nan, 8., nan, 5., 3., 7.])
b
# array([nan, 2., 3., nan, 7., nan, 6., nan, 6., 9.])
我知道这与 OP 的“预期输出”不同,但根据他们的解释,我认为他们的 output 中有一个错误。
if len(a) > len(b):
for i, element in enumerate(b):
if np.isnan(element) or np.isnan(a[i]):
a[i] = np.nan
b[i] = np.nan
else:
for i, element in enumerate(a):
if np.isnan(element) or np.isnan(b[i]):
a[i] = np.nan
b[i] = np.nan
这是一个解决方案:
for i in range(min(len(a), len(b))):
if np.isnan(a[i]) or np.isnan(b[i]):
a[i] = np.NaN
b[i] = np.NaN
print(a)
# [nan, 2, 3, nan, 4, nan, 8, nan, 5, 3, 7]
print(b)
# [nan, 2, 3, nan, 7, nan, 6, nan, 6, 9]
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