返回列表中最接近的值

正在输出的列表如下。我想要做的是找到第二列中的数字最接近的值加上一个步长。我会如何去做这件事？我曾尝试使用min函数，但它不可迭代。返回列表中最接近的值

stepsize = .5 
return closest(column[1] + stepsize) 

(72817, 10.009872745252624, 40.999890710175876, 9.73) 
(103394, 10.044319950550072, 11.450070211613395, 8.1) 
(52251, 10.047512210212679, 73.31363177934391, 9.75) 
(98118, 10.558521350586966, 24.665802379879878, 9.13) 
(101401, 10.595011636219427, 17.691450116195412, 8.34) 
(90376, 10.718434008267023, 32.37843662097162, 9.24) 
(53624, 10.78156459297739, 65.73007957579946, 8.43) 
(99855, 10.960898039784297, 20.83812851138556, 9.05) 
(91937, 10.999664485957146, 31.048708072900475, 9.04) 
(97166, 11.049670008406684, 25.46563972962055, 9.92)

所以它应该返回10.558521 ...这个列表会持续很长时间，所以步长会有所不同。

来源

2014-08-27 Greg J

怎么会加入0.5有什么区别，如果你把它添加到所有的人？ – 2014-08-27 17:16:00

我不想将它添加到所有这些，我想找到第一个元素+ stepsize的最接近的值，然后对于下一个它将找到最接近该数字+ stepize再次。对不起，如果我没有说清楚 – 2014-08-27 17:20:51

stepsize = 0.5 
grid = [(72817, 10.009872745252624, 40.999890710175876, 9.73), 
     (103394, 10.044319950550072, 11.450070211613395, 8.1), 
     (52251, 10.047512210212679, 73.31363177934391, 9.75), 
     (98118, 10.558521350586966, 24.665802379879878, 9.13), 
     (101401, 10.595011636219427, 17.691450116195412, 8.34), 
     (90376, 10.718434008267023, 32.37843662097162, 9.24), 
     (53624, 10.78156459297739, 65.73007957579946, 8.43), 
     (99855, 10.960898039784297, 20.83812851138556, 9.05), 
     (91937, 10.999664485957146, 31.048708072900475, 9.04), 
     (97166, 11.049670008406684, 25.46563972962055, 9.92)] 

def closest(value, data): 
    """Returns the closest element in data to the value passed""" 
    return min(data, key=lambda x: abs(x-value)) 

results = [(row[1], closest(row[1]+stepsize, map(lambda x: x[1], grid))) for row in grid] 
# or simplified using Padraic's zip method to grab the relevant column: 
## data = (zip(*grid))[1] 
## results = [(datum, closest(datum+stepsize, data)) for datum in data]

结果：

results = [(10.009872745252624, 10.558521350586966), 
      (10.044319950550072, 10.558521350586966), 
      (10.047512210212679, 10.558521350586966), 
      (10.558521350586966, 11.049670008406684), 
      (10.595011636219427, 11.049670008406684), 
      (10.718434008267023, 11.049670008406684), 
      (10.78156459297739, 11.049670008406684), 
      (10.960898039784297, 11.049670008406684), 
      (10.999664485957146, 11.049670008406684), 
      (11.049670008406684, 11.049670008406684)]

来源

2014-08-27 17:25:53

排序每个元素的abs差得到的只是使用zip然后第一列x + .5-你想让它最接近号码：

col = (zip(*l))[1] # l is your list of tuples 

print sorted(col,key=lambda x: abs(x + 0.5 - 10.6))` 
[10.047512210212679, 10.044319950550072, 10.009872745252624,10.558521350586966,10.595011636219427, 10.718434008267023, 10.78156459297739, 10.960898039784297, 10.999664485957146, 11.049670008406684]

sorted(col,key=lambda x: abs(x + 0.5 - 10.6))[0]将是最接近

为了把它放在一个功能：

def closest(l,step,val): 
    col = (zip(*l))[1] 
    return sorted(col,key=lambda x: abs(x + step - val))[0]

来源

2014-08-27 17:21:20

返回列表中最接近的值

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