Python：对于每个列表元素，在列表中应用一个函数

Question

Given [1,2,3,4,5] , how can I do something like

1/1, 1/2, 1/3,1/4,1/5, ...., 3/1,3/2,3/3,3/4,3/5,.... 5/1,5/2,5/3,5/4,5/5

I would like to store all the results, find the minimum, and return the two numbers used to find the minimum. So in the case I've described above I would like to return (1,5) .

So basically I would like to do something like

for each element i in the list map some function across all elements in the list, taking i and j as parameters store the result in a master list, find the minimum value in the master list, and return the arguments i , j used to calculate this minimum value.

In my real problem I have a list objects/coordinates, and the function I am using takes two coordinates and calculates the euclidean distance. I'm trying to find minimum euclidean distance between any two points but I don't need a fancy algorithm.

Answer 1

You can do this using list comprehensions and min() (Python 3.0 code):

>>> nums = [1,2,3,4,5]
>>> [(x,y) for x in nums for y in nums]
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)]
>>> min(_, key=lambda pair: pair[0]/pair[1])
(1, 5)

Note that to run this on Python 2.5 you'll need to either make one of the arguments a float, or do from __future__ import division so that 1/5 correctly equals 0.2 instead of 0.

Answer 2

If I'm correct in thinking that you want to find the minimum value of a function for all possible pairs of 2 elements from a list...

l = [1,2,3,4,5]

def f(i,j):
   return i+j 

# Prints min value of f(i,j) along with i and j
print min( (f(i,j),i,j) for i in l for j in l)

Answer 3

Some readable python:

def JoeCalimar(l):
    masterList = []
    for i in l:
        for j in l:
            masterList.append(1.*i/j)
    pos = masterList.index(min(masterList))
    a = pos/len(masterList)
    b = pos%len(masterList)
    return (l[a],l[b])

Let me know if something is not clear.

Answer 4

If you don't mind importing the numpy package, it has a lot of convenient functionality built in. It's likely to be much more efficient to use their data structures than lists of lists, etc.

from __future__ import division

import numpy

data = numpy.asarray([1,2,3,4,5])
dists = data.reshape((1,5)) / data.reshape((5,1))

print dists

which = dists.argmin()
(r,c) = (which // 5, which % 5) # assumes C ordering

# pick whichever is most appropriate for you...
minval = dists[r,c]
minval = dists.min()
minval = dists.ravel()[which]

Answer 5

Doing it the mathy way...

nums = [1, 2, 3, 4, 5]
min_combo = (min(nums), max(nums))

Unless, of course, you have negatives in there. In that case, this won't work because you actually want the min and max absolute values - the numerator should be close to zero, and the denominator far from it, in either direction. And double negatives would break it.

Answer 6

If working with Python ≥2.6 (including 3.x), you can:

from __future__ import division
import operator, itertools

def getmin(alist):
    return min(
        (operator.div(*pair), pair)
        for pair in itertools.product(alist, repeat=2)
    )[1]

getmin([1, 2, 3, 4, 5])

EDIT: Now that I think of it and if I remember my mathematics correctly, this should also give the answer assuming that all numbers are non-negative:

def getmin(alist):
    return min(alist), max(alist)

Answer 7

>>> nums = [1, 2, 3, 4, 5]    
>>> min(map((lambda t: ((float(t[0])/t[1]), t)), ((x, y) for x in nums for y in nums)))[1]
(1, 5)