如何在 python 中对 csv 文件中的范围进行排序？

Question

我的问题：我有一个 csv 文件，其中包含 90 到 3 米深的数据，并且这些数据会像这样来回传输。我正在使用最新的python。

例如。 (深度) 88, 77, 50, 20, 5, 90, 76, 54, 34, 15, 8, 4, 81, 74, 62,51, 49, 30, 22, 10, 8...等.它一直从 90 到 3，然后又回来。

我想要做的是每次数据介于 90 和 3 之间时将其分开。一旦分开，我想获取该列表中的最后一个值和第一个值。 像这样 前任。 88, 77, 50, 20, 5 (这里分开), 90, 76, 54, 34, 15, 8, 4 (这里分开) 81, 74, 62,51, 49, 30, 22, 10, 8 这里分开）... 等等。它一直从 90 到 3，然后又回来。

我该怎么做呢？或者我如何将它们分成列表，然后使用每个列表中的数据？

这是我目前的代码：

import csv, numpy
from collections import defaultdict

columns = defaultdict(list) # each value in each column is appended to a list

with open('C:\\Users\\AdamStoer\\Documents\\practicedata.csv') as f:

reader = csv.DictReader(f,delimiter=',') # read rows into a dictionary format
for row in reader:
    r = float(row['roll'])
    p = float(row['pitch'])
    if 0.21 <= p <= 0.31:
            if -0.06 <= r <= 0.06:
                columns['pitch'].append(row['pitch'])
                columns['roll'].append(row['roll'])
                columns['i_depth'].append(row['i_depth'])
                columns['irrad2'].append(row['sci_ocr504i_irrad2'])

print ('Pitch:')
print (columns['pitch'])
print ('Roll:')
print (columns['roll'])
print ('Depth')
print (columns['i_depth'])
print ("Irrandiance(2):")
print (columns['irrad2'])


irradlst = columns['irrad2']
irradfirst = irradlst[0]
irradlast = irradlst[-1]

depthlst = columns['i_depth']
depthfirst = depthlst[0]
depthlast = depthlst[-1]

print ("\nDepth 1 is " + depthfirst + " and " + "Depth 2 is " + depthlast)

print ("\nIrradiance 1 is " + irradfirst + " and " + "Irradiance 2 is " +     irradlast)

#Find the Volume Attenuation Coefficient

#irranddiv =  deepest/shallowest
irraddiv = float(irradfirst)/float(irradlast)

#depthdif = deepest-shallowest
depthdif = float(depthfirst) - float(depthlast)

#Find Log of irraddiv
irradlog = numpy.log(irraddiv)           

#Find K
K = irradlog/(-depthdif)

print("\nAttenuation Coefficient")
print (K)

Answer 1

好吧，你的代码有点复杂，我不知道 numpy，无论如何这是我想出的用于分隔数字范围的解决方案：

l = [88, 77, 50, 20, 5, 90, 76, 54, 34, 15, 8, 4, 81, 74, 62,51, 49, 30, 22, 10, 8,65]

group =0 #since were using dictionaries i use group number as dictionary KEY to distinguish each group of number between 3 to 90 
temp = [] # this is a list that we keep temporary group of our number ranged between 3 to 90
splited_list = {} #this is a dictionary that we keep our final result in side it
lengh = len(l) #this is a lengh of our list of numbers

for i in range(lengh): # we run our code for each number inside our list of number
 if not i == lengh-1: # at this line we check if our number is not the last number in list , because in next line we check our number with the number that comes after our current number and if it's the last number we get "IndexError: list index out of range" error at next line
  if l[i] > l[i+1]: # since our range is between 3 to 90 , so if our current number is bigger than next number, for sure it is bigger than 3, so it is between 3 to 90 and we can add it to our current group of number
   temp.append(l[i])
  else: #if it is not bigger than the next number it is our last number in our current group of number in range of 90 to 3 so we add it to our temp 
   temp.append(l[i])
   group +=1
   splited_list.update({str(group):temp})
   temp = []
 else: # when we reach this line it means that we get to the last number in ourlist , since there is no next number , we check it with previous number , if its bigger it is not belong to current group of number between 3 to 90 and if it's smaller it is belong to the current group of number  
  if l[i] < l[-2]:
   temp.append(l[i])
   group +=1
   splited_list.update({str(group):temp})
   break
  else:
   group +=1
   splited_list.update({str(group):[l[i]]})
   break

它根据需要将范围 90 到 3 分开，它给出以下输出：

>>> splited_list
{'2': [90, 76, 54, 34, 15, 8, 4], '1': [88, 77, 50, 20, 5], '4': [65], '3': [81, 74, 62, 51, 49, 30, 22, 10, 8]}

Answer 2

这个任务很简单，numpy.diff() 和 numpy.where() 如下：

代码：

import numpy as np

def split_at_mins(a_list):
    end_points = list(1 + np.where(0 < np.diff(a_list))[0])
    return [a_list[i:j] for i, j in
            zip([0] + end_points, end_points + [len(a_list)])]

测试代码：

test_data = (
    88, 77, 50, 20, 5,
    90, 76, 54, 34, 15, 8, 4,
    81, 74, 62, 51, 49, 30, 22, 10, 8
)

print('\n'.join(str(d) for d in split_at_mins(test_data)))
print('\n'.join('%s %s' % (d[0], d[-1]) for d in split_at_mins(depth)))

生产：

(88, 77, 50, 20, 5)
(90, 76, 54, 34, 15, 8, 4)
(81, 74, 62, 51, 49, 30, 22, 10, 8)
88 5
90 4
81 8