Python，将2d列表中的所有数字除以10，然后返回2d列表

Question

I wrote this solution, and coming from Ruby it seems very elegant. However, is this the way a python programmer would do it?

a = [[2,3,4], [9,1,2]]
print map(lambda(i): map(lambda(p): p/10.0,i), a)

And... what if instead of 10, I wanted to use the total of all the values in the nested 2d list?

Answer 1

That's generally solved by using comprehensions, in this case a nested list-comprehension:

>>> from __future__ import division   # for python-2.x compatibility
>>> [[item / 10 for item in subl] for subl in a]
[[0.2, 0.3, 0.4], [0.9, 0.1, 0.2]]

That's probably faster than map and avoids all the lambda functions.

what if instead of 10, I wanted to use the total of all the values in the nested 2d list?

Calculate the total using sum and a nested generator expression:

>>> sum_ = sum(item for subl in a for item in subl)
>>> [[item / sum_ for item in subl] for subl in a]
[[0.09523809523809523, 0.14285714285714285, 0.19047619047619047],
 [0.42857142857142855, 0.047619047619047616, 0.09523809523809523]]

---

But with NumPy arrays it's even easier. NumPy a 3rd party package but very powerful and fast:

>>> import numpy as np
>>> arr = np.array(a)
>>> arr / 10.   # element-wise division
array([[ 0.2,  0.3,  0.4],
       [ 0.9,  0.1,  0.2]])

>>> arr / arr.sum()  # sum over all elements then element-wise division
array([[ 0.0952381 ,  0.14285714,  0.19047619],
       [ 0.42857143,  0.04761905,  0.0952381 ]])

Answer 2

Another approach is to use numpy , since it deals with arrays and its operators work on each element of the array by default.

In [1]: import numpy as np

In [2]: a = np.array([[2,3,4], [9,1,2]])

In [3]: a/10
Out[3]: 
array([[ 0.2,  0.3,  0.4],
       [ 0.9,  0.1,  0.2]])

To divide by the sum of all the numbers, do

In [6]: a/a.sum()
Out[6]: 
array([[ 0.0952381 ,  0.14285714,  0.19047619],
       [ 0.42857143,  0.04761905,  0.0952381 ]])

Answer 3

List comprehensions are much more elegant and would be considered more Pythonic (with their better readability):

>>> [[x/10.0 for x in lst] for lst in a]
[[0.2, 0.3, 0.4], [0.9, 0.1, 0.2]]

Also, note that map won't return a list in Python 3, it returns an iterator object. You'll need to call list on each of the map objects to have them evaluated into list. The list comp returns the same result across versions.