Python 3.6.3中用map处理numpy数组遇TypeError问题求助

Let's break down what's happening in your code and how to fix it:

Why the Error Occurs #

Your seen variable is a 2D numpy array (shape (3,1)):

seen = np.array([[0,1,0]]).T
# Output: array([[0],
#                [1],
#                [0]])

When map iterates over seen, each x passed to the lambda is a 1-element numpy array (like array([0])), not a plain integer. Python lists (like observations) only accept integer scalars as indices—you can't use an array to index into a list, hence the error.

How to Fix It #

You have a few straightforward solutions:

1. Flatten the seen Array #

Convert the 2D array to a 1D array so each iteration gives an integer scalar:

import numpy as np

observations = ["red", "white"]
seen = np.array([[0,1,0]]).T

# Use flatten() to get a 1D array
result = list(map(lambda x: observations[x], seen.flatten()))
print(result)  # Output: ['red', 'white', 'red']

2. Extract the Scalar from Each Array Element #

Modify your lambda to pull the integer value out of the 1-element array:

result = list(map(lambda x: observations[x[0]], seen))
print(result)  # Output: ['red', 'white', 'red']

3. Use Numpy's Built-in Indexing (More Idiomatic) #

Since you're already using numpy, skip map entirely and use numpy's native array indexing:

# Convert observations to a numpy array first, then index
numpy_obs = np.array(observations)
result = numpy_obs[seen.flatten()].tolist()
# Or directly handle the 2D array:
result = numpy_obs[seen].flatten().tolist()
print(result)  # Output: ['red', 'white', 'red']

How to View Map Results #

The map object you see (<map at 0x2ab8c869e80>) is an iterator—it doesn't compute results until you iterate over it. To see the actual values, convert it to a list with list(map(...)) (which is what you were trying to do, but the error stopped you). Once you fix the indexing issue, this will work as expected.

内容的提问来源于stack exchange，提问作者Sunny