Python 对于大小为 0 的轴 0，索引 0 超出范围
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index 0 is out of bounds for axis 0 with size 0
提问by johnhenry
I am filling two arrays, field_in_k_space_REALand field_in_k_space_IMAGINARY, with values extracted from a Gaussian distribution, paying attention to respect the symmetry to get a real field when I inverse-transform the arrays. Here is the code:
我正在填充两个数组，field_in_k_space_REAL并field_in_k_space_IMAGINARY使用从高斯分布中提取的值，注意在对数组进行逆变换时尊重对称性以获得实场。这是代码：
field_in_k_space_REAL = zeros(n, float)
field_in_k_space_IMAGINARY = zeros(n, float)
field_in_k_space_REAL[0] = 0.0
for i in range(1, int(n/2+1)):
field_in_k_space_REAL[i] = np.random.normal(mu, math.sqrt((1/2)*math.exp(-(2*math.pi*i*sigma/L)*(2*math.pi*i*sigma/L))))
x = range(int(n/2+1), int(n))
y = range(1, int(n/2))
zipped = zip(x, y)
for j, j2 in zipped:
field_in_k_space_REAL[j] = field_in_k_space_REAL[j-2*j2]
field_in_k_space_IMAGINARY[0] = 0.0
for i in range(1, int(n/2)):
field_in_k_space_IMAGINARY[i] = np.random.normal(mu, math.sqrt((1/2)*math.exp(-(2*math.pi*i*sigma/L)*(2*math.pi*i*sigma/L))))
field_in_k_space_IMAGINARY[n/2] = 0.0
for j, j2 in zipped:
field_in_k_space_IMAGINARY[j] = - field_in_k_space_IMAGINARY[j-2*j2]
print 'field_k', field_in_k_space_REAL
But I keep having the following error:
但我一直有以下错误：
field_in_k_space_REAL[0] = 0.0
IndexError: index 0 is out of bounds for axis 0 with size 0
Can someone explain why and how to fix it?
有人可以解释为什么以及如何解决它吗？
采纳答案by James Kelleher
My guess is that the array field_in_k_space_REALis actually of length 0, most likely because you set n = 0further up in your code (do you use nin a loop maybe?). I can reproduce the error when I directly initialize an array of length 0.
我的猜测是该数组field_in_k_space_REAL实际上长度为 0，很可能是因为您n = 0在代码中进行了进一步设置（您是否n在循环中使用？）。当我直接初始化长度为 0 的数组时，我可以重现该错误。
回答by Jansora
in fact, you'd better use:
事实上，你最好使用：
field_in_k_space_REAL.loc[index] = 0.0
instead of:
代替：
field_in_k_space_REAL[index] = 0.0
回答by Akshay Salvi
this is because of NAN value in your close column use np.isfinite. For example :
这是因为您的关闭列中的 NAN 值使用np.isfinite。例如 ：
btt_ohlc[np.isfinite(btt_ohlc['Close'])]
