英文:
Print time at which minimum distance occurs given a set of reference distance values
问题
从列表A中的一组值中，我想知道与列表B中的值的最小距离发生的时间（第一列），这些最小距离在第二列中报告。
import numpy as np
list_A = np.loadtxt('listA.dat')
file_B = np.loadtxt('listB.dat')
distances = file_B[:,1]
time = file_B[:,0]
abs_differences = []
for point in list_A:
diffs = []
for distance in distances:
diffs.append(abs(distance - point))
min_index = np.argmin(diffs)
min_time = time[min_index]
abs_differences.append(min_time)
with open("output.txt", "w") as f:
for min_time in abs_differences:
print(min_time, file=f)
然而，输出文件需要很长时间才能写入，因为对于数据集中的每个点，都会打印出列表B中的每个点发生最小距离的时间，因此会有多次出现时间值。是否有解决这个问题的方法，可能可以简化代码？
请注意，我已经简化了代码并修复了其中的错误。现在，它将仅打印每个点的最小时间值到输出文件中，而不是多次打印。
英文:
From a list of values in list A, I want to know the time (first column) for a list B in which the minimum distances with values of the list A occur (distances reported in second column).
import numpy
list_A = np.loadtxt('listA.dat')
file_B = np.loadtxt('listB.dat')
distances = file_B[:,1]
time = file_B[:,0]
abs_differences = []
for point in list_A:
diffs = []
for distance in distances:
diffs.append(abs(distance-point))
list={}
list=['Point'] = point
list=['AbsDiff'] = diffs
abs_differences.append(list)
for ele in abs_differences:
diffs = ele['AbsDiff']
index = diffs.index(min(diffs))
with open("output.txt","a") as f:
print(time[index], file=f)
However, the output file takes a really long time to be written, since the time at which the minimum distance occurs for every point of list B is printed for every point of the dataset, therefore having multiple occurrences of the time values. Is there a way to solve this issue, possibly simplifying the code?
答案1
得分: 1
您在代码中多次切换类型，并将Python关键字list用作变量名，因此至少出现了这两个原因，您的代码无法正常工作。
您可以尝试基于以下简化的代码：
import numpy as np
list_A = np.loadtxt('listA.dat')
file_B = np.loadtxt('listB.dat')
list_B = file_B[:, 1]
times = file_B[:, 0]
min_times = [] # 元组列表（pointA，pointB，indexB，distance，time）
for pointA in list_A:
min_distance = float('inf')
min_time = None
for indexB, pointB in enumerate(list_B):
distance = abs(pointA - pointB)
if distance < min_distance:
min_distance = distance
min_time = times[indexB]
min_times.append((pointA, pointB, indexB, min_distance, min_time))
print(min_times)
# 如果您想要距离：（元组中的索引3）
print([mt[3] for mt in min_times])
# 一旦确定所需的结果，您还可以将其写入文件
请注意，上述代码已经根据您提供的简化版本进行了翻译。
英文:
You are switching between types several times and using the python keyword list as a variable name, so your code cannot work properly at least for these two reasons.
You should try something based on the following simplified code:
import numpy as np
list_A = np.loadtxt('listA.dat')
file_B = np.loadtxt('listB.dat')
list_B = file_B[:,1]
times = file_B[:,0]
min_times = [] # list of tuples (pointA, pointB, indexB, distance, time)
for pointA in list_A:
min_distance = float('inf')
min_time = None
for indexB, pointB in enumerate(list_B):
distance = abs(pointA - pointB)
if distance < min_distance:
min_distance = distance
min_time = times[indexB]
min_times.append((pointA, pointB, indexB, min_distance, min_time))
print(min_times)
# if you want the distances: (index 3 in the tuple)
print([mt[3] for mt in min_times])
# you can also write to file once you are sure of the result wanted
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论