如何將點打印爲數組？

Question

如何改進以下適用於數組列表的公式。我不想手動輸入x和y，並且我想知道這個距離與哪個點有關

例如：dis: (2,4) & (-4,2) = ?? dis: (-4,2) & (10,-5) = ??

import math 
array = [(2, 4), (-4, 2),(10, -5)] 


def Distance(x1, y1, x2, y2): 

    dist= math.sqrt((x2-x1)**2 +(y2-y1)**2) 
    return dist 

Distance(array) 

Answer 1

這將返回一個包含輸入爲元組（X，Y）每兩個連續點之間的（歐幾里得）距離的列表。這些代碼片段假設Python 3.x，如果您使用Python 2.7，則可能需要進行小的更改。您的標籤不清楚這一點。

import math 


def distances(lst): 
    ln = len(lst) 
    dists = [] 
    if ln >= 2: # if array contains no or only 1 tuple, return [] 
     for index, tup in enumerate(lst[:ln - 1]): 
      x1, y1 = tup 
      x2, y2 = lst[index + 1] # gets x and y of next tuple in the array 
      dists.append(math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2)) 
    return dists 


array0 = [] 
array1 = [(2, 4), (-4, 2)] 
array2 = [(5, 2), (-1, 3), (6, -6)] 
array3 = [(2, 4), (-4, 2), (10, -5), (9, -2)] 

print(distances(array0)) 
print(distances(array1)) 
print(distances(array2)) 
print(distances(array3)) 

輸出：

[] 
[6.324555320336759] 
[6.082762530298219, 11.40175425099138] 
[6.324555320336759, 15.652475842498529, 3.1622776601683795] 

如果你要打印的距離，你提到的方式：

import math 


def distances(lst): 
    ln = len(lst) 
    if ln >= 2: 
     for index, tup in enumerate(lst[:ln - 1]): 
      x1, y1 = tup 
      x2, y2 = lst[index + 1] 
      dist = math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2) 
      print(lst[index:index + 2], "=", dist) 


array = [(5, 2), (-1, 3), (6, -6)] 

distances(array) 

輸出：

[(5, 2), (-1, 3)] = 6.082762530298219 
[(-1, 3), (6, -6)] = 11.40175425099138 

相同的功能之前可以使用numpy模塊instea來實現數學的d：

import numpy as np 


def distances(lst): 
    ln = len(lst) 
    dists = [] 
    if ln >= 2: 
     for index, tup in enumerate(lst[:ln - 1]): 
      dists.append(np.linalg.norm(lst[index + 1] - tup)) 
    return dists 


array0 = np.array([]) 
array1 = np.array([(2, 4), (-4, 2)]) 
array2 = np.array([(5, 2), (-1, 3), (6, -6)]) 
array3 = np.array([(2, 4), (-4, 2), (10, -5), (9, -2)]) 

print(distances(array0)) 
print(distances(array1)) 
print(distances(array2)) 
print(distances(array3)) 

輸出：

[] 
[6.324555320336759] 
[6.0827625302982193, 11.401754250991379] 
[6.324555320336759, 15.652475842498529, 3.1622776601683795] 

如果你只想要包含在列表中有兩個元組之間的距離：

import math 


def distance(lst): 
    return math.sqrt((lst[1][0] - lst[0][0]) ** 2 + (lst[1][1] - lst[ 
     0][1]) ** 2) 


array = [(2, 4), (-4, 2)] 

print(distance(array)) 

輸出：

6.324555320336759 

Answer 2

假設你使用Python 2.7，因爲它不清楚：

import math 


def Distance(x1, y1, x2, y2): 
    return math.sqrt((x2-x1)**2 +(y2-y1)**2) 

points = [(5, 2), (-1, 3), (6, -6)] 

pairs = [(points[i], points[i+1]) for i in range(len(points)-1)] 

for val in pairs: 
    print 'Distance between {0} and {1}: {2}'.format(val[0], val[1], Distance(*val[0]+val[1])) 

我們用一個列表理解制定出每對點（假設你只需要在列表中每對連續值），然後用元組拆包養活這些作爲輸入到Distance方法。

如果您在Python 3.x中，最後一行應該是：

print('Distance between {0} and {1}: {2}'.format(val[0], val[1], Distance(*val[0], *val[1]))) 

有了您給出的例子，這個打印：

Distance between (5, 2) and (-1, 3): 6.0827625303 
Distance between (-1, 3) and (6, -6): 11.401754251