如何使用 isclose() 将坐标列表中的重复项数作为元组获取？

Question

我有一个元组坐标列表：

  listoc = ['(-87.5, -960.8147117258)', 
     '(91.3147117258, -960.8147117258)', 
     '(85.5, 99.50000000000011)', 
     '(85.5, -966.8147117258)', 
     '(91.3147117258, -968.8147117258)', 
     '(87.5, -968.8147117258)', 
     '(93.3147117258, -966.8147145072)', 
     '(93.3147117258, -962.8147117258)', 
     '(-87.5, -968.8147117258)', 
     '(-91.3147117258, -155.00000000000009)', 
     '(-85.5, -966.8147145204)', 
     '(-85.5, -962.8147117258)', 
     '(-91.3147117258, -155.0)',
     '(-87.5, -960.8147117258)', 
     '(-93.3147117258, -962.8147089444)', 
     '(85.500000071, 99.50)', 
     '(-73.0, -517.5)']

随着math.isclose()的使用我想得到重复的个数

我在没有 isclose() 的情况下尝试过这个，因为我不知道如何:

    for cord in listoc:
        number_cc += listoc.count(cord)

    print('Number of duplicates: '+ str(number_cc))

输出：

Number of duplicates: 19

但应该是：

Number of duplicates: 3

我正在使用 python3.6

Answer 1

使用：

for cord in listoc:
        number_cc += listoc.count(cord) - 1  ### here the difference

print('Number of duplicates: '+ str(number_cc/2)) ### here the difference

否则你至少会得到 listoc 的元素数量，因为 cord 它总是至少在列表中出现一次，并且你计算了两次相同的数字

根据你的标题，如果你想使用 math.isclose()，我最终得到：

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sun Jun 20 12:34:52 2021

@author: Pietro

https://whosebug.com/questions/68054705/how-to-get-the-number-of-duplicates-in-a-list-of-coordinates-as-tuples-with-iscl

"""

import math

import itertools



listoc = ['(-87.5, -960.8147117258)', 
     '(91.3147117258, -960.8147117258)', 
     '(85.5, 99.50000000000011)', 
     '(85.5, -966.8147117258)', 
     '(91.3147117258, -968.8147117258)', 
     '(87.5, -968.8147117258)', 
     '(93.3147117258, -966.8147145072)', 
     '(93.3147117258, -962.8147117258)', 
     '(-87.5, -968.8147117258)', 
     '(-91.3147117258, -155.00000000000009)', 
     '(-85.5, -966.8147145204)', 
     '(-85.5, -962.8147117258)', 
     '(-91.3147117258, -155.0)',
     '(-87.5, -960.8147117258)', 
     '(-93.3147117258, -962.8147089444)', 
     '(85.500000071, 99.50)', 
     '(-73.0, -517.5)']

number_cc = 0
for cord, cord2 in itertools.combinations(listoc, 2):
    

        cordx, cordy = cord.split(',')[0].strip('('), cord.split(',')[1].strip(')')
        cord2x, cord2y = cord2.split(',')[0].strip('('), cord2.split(',')[1].strip(')')
        if math.isclose(float(cordx),float(cord2x)) and math.isclose(float(cordy),float(cord2y)):
            
            number_cc +=1  
            print('ok #: ' , number_cc,' index : ',listoc.index(cord),cord,' index : ',  listoc.index(cord2),cord2)
        else:
            pass
            
print(number_cc)

输出为：

ok #:  1  index :  0 (-87.5, -960.8147117258)  index :  0 (-87.5, -960.8147117258)
ok #:  2  index :  2 (85.5, 99.50000000000011)  index :  15 (85.500000071, 99.50)
ok #:  3  index :  9 (-91.3147117258, -155.00000000000009)  index :  12 (-91.3147117258, -155.0)
3

所以 3 个重复，要小心，因为这里：

ok #:  1  index :  0 (-87.5, -960.8147117258)  index :  0 (-87.5, -960.8147117258)

第二个索引 0 应该是 13 但是 list.index(x) return 第一个索引有 x 作为值，它实际上是你列表中唯一的 x == y

你可以玩：

math.isclose(a, b, rel_tol, abs_tol)

其中：

a   Required. The first value to check for closeness

b   Required. The second value to check for closeness

rel_tol = value     Optional. The relative tolerance. It is the maximum allowed difference between value a and b. Default value is 1e-09

abs_tol = value     Optional. The minimum absolute tolerance. It is used to compare values near 0. The value must be at least 0

Answer 2

import collections

listoc = ['(-87.5, -960.8147117258)',
 '(91.3147117258, -960.8147117258)',
 '(85.5, 99.50000000000011)',
 '(85.5, -966.8147117258)',
 '(91.3147117258, -968.8147117258)',
 '(87.5, -968.8147117258)',
 '(93.3147117258, -966.8147145072)',
 '(93.3147117258, -962.8147117258)',
 '(-87.5, -968.8147117258)',
 '(-91.3147117258, -155.00000000000009)',
 '(-85.5, -966.8147145204)',
 '(-85.5, -962.8147117258)',
 '(-91.3147117258, -155.0)',
 '(-87.5, -960.8147117258)',
 '(-93.3147117258, -962.8147089444)',
 '(85.500000071, 99.50)',
 '(-73.0, -517.5)']

print([(item, count) for item, count in collections.Counter(listoc).items() if count > 1])

如何使用 isclose() 将坐标列表中的重复项数作为元组获取？

How to get the number of duplicates in a list of coordinates as tuples with isclose()?

python

tuples

list

python-3.x