'int' object 不可迭代，map() 通过列表传递

Question

So lets I have this list;

yy=[1,2,3,4,5,6,7,8,9,10]

and I want a map() to go through the list using the np.std() this,

np. std(1,2,3), std(2,3,4), std(3,4,5).... std(8,9,10)

so I thought of doing something like this,

import numpy as np
y = [1,2,3,4,5,6,7,8,9,10]
n = 3
x = map(lambda w, n: np.std(y[x-n:x]), range(n,len(y)), n)
print(list(x))

But I get this error 'int' object is not iterable, how do I fix this?

Answer 1

You're passing too many iterables to map . You just need to pass the start index for each piece of the list that you want to np.std()

import numpy as np
y = [1,2,3,4,5,6,7,8,9,10]
n = 3
x = map(lambda i: np.std(y[i:i+n]), range(len(y) - n)) # pass only the start index for each iteration
print(list(x))

Result:

[0.816496580927726, 0.816496580927726, 0.816496580927726, 0.816496580927726, 0.816496580927726, 0.816496580927726, 0.816496580927726]

Answer 2

OK, now I see what you were trying to do. You expected the single value of n to be passed each time into your lambda function. But that's not how map works, it expects all its arguments to be iterables.

You had another problem, you are passing w into your lambda function but trying to use it as x .

Here's the version of your code with those two problems fixed:

x = map(lambda w, n: np.std(y[w-n:w]), range(n,len(y)), [n]*len(y))

You can simplify it by realizing that you don't need to pass n into the lambda function at all, it's already defined in the scope where you're creating the function.

x = map(lambda w: np.std(y[w-n:w]), range(n,len(y)))

Finally you can use a generator expression instead of map .

x = (np.std(y[w-n:w]) for w in range(n,len(y)))