Question

我在text.txt文件中有一个数字列表 2.50
2.56
2.81
2.86
2.84
3.21
3.47
2.91
2.96
3.11
2.83
2.89
2.94
2.94
3.34
3.44
2.94
2.96
3.04
3.01
2.85
3.05
3.10

我希望将每组数量的范围加起来。比如一个范围内有多少 2.5-2.7
2.7-2.9
2.9-3.1
3.1-3.3
3.3-3.5

我试过这个。

from __future__ import division
from math import *
from numpy import *
from string import*

infile = open('text1.txt', 'r')
text = infile.read().split('\n')
infile.close()
text.remove('')

numbers = []
for i in text:
count = 0
if (numbers[i] > 2.49) and (numbers[i] < 2.59):
    count += 1
    print("Number of elements", count)

它不能正常工作

Answer 1

您可以使用bisect模块：

>>> import bisect
>>> ranges = [2.5, 2.7, 2.9, 3.1, 3.3, 3.5]
>>> nums = [2.5, 2.56, 2.81, 2.86, 2.84, 3.21, 3.47, 2.91, 2.96, 3.11, 2.83, 2.89, 2.94, 2.94, 3.34, 3.44, 2.94, 2.96, 3.04, 3.01, 2.85, 3.05, 3.1]
>>> lis = [0]*len(ranges)
for item in nums:
    ind = bisect.bisect(ranges, item) - 1
   lis[ind] += 1
for x, y in zip(zip(ranges, ranges[1:]), lis):
   print x, y
...     
(2.5, 2.7) 2
(2.7, 2.9) 6
(2.9, 3.1) 9
(3.1, 3.3) 3
(3.3, 3.5) 3

Answer 2

如何使用更多numpy函数？

import numpy

numbers = numpy.loadtxt('test.txt')
bins = numpy.arange(2.5, 3.51, 0.2) #  3.5 won't work due to floating point issues
counts, _ = numpy.histogram(numbers, bins)

如果您不想使用numpy，您可以通过直接计算相等大小的二进制数字落入哪个bin来获益：

numbers = [float(n) for n in open('test.txt') if len(n.strip())]
start = 2.5
width = 0.2
end = 3.7

def position(n):
    return int((n - start)/width)

counts = [0 for i in range(position(end))]
for n in numbers:
    counts[position(n)] += 1

Answer 3

这不起作用，因为你没有存储数字[]。

numbers = []
count = 0
for i in text:
    numbers.append(int(i))
    count=count+1

count = 0
for i in text:
    if (numbers[i] > 2.49) and (numbers[i] < 2.59):
        count += 1
print("Number of elements", count)

Answer 4

首先，您可以使用readlines()来改善文件阅读：

numbers = [float(i.strip()) for i in infile.readlines() if i is not '']

接下来，对于bin计数，假设每个bin的范围相等，您可以创建两个指定起始值和delta的变量：

start = 2.5
delta = 0.2
nBins = 5

然后您可以使用filter来获取每个范围的计数：

counts = [len(filter(lambda x: start+delta*i <= x < start+delta*(i+1), numbers)) for i in xrange(nBins)]

并打印结果：

for i,count in enumerate(counts):
    print "Number of elements in the range %.1f-%.1f: %d" % (start+delta*i,start+delta*(i+1),count)

完整代码：

infile = open('text1.txt', 'r')
numbers = [float(i.strip()) for i in infile.readlines() if i is not '']

start = 2.5
delta = 0.2
nBins = 5

counts = [len(filter(lambda x: start+delta*i <= x < start+delta*(i+1), numbers)) for i in xrange(nBins)]

for i,count in enumerate(counts):
    print "Number of elements in the range %.1f-%.1f: %d" % (start+delta*i,start+delta*(i+1),count)

Answer 5

试试这个：

ranges = [(2.5, 2.7), (2.7, 2.9), (2.9, 3.1), (3.1, 3.3), (3.3, 3.5)]
counts = {i:0 for i in ranges}

def findBin(n, bins):
    mid = len(bins)/2
    low, high = bins[mid]
    if low <= n <= high:
        return (low,high)
    elif low >= n:
        return findBin(n, bins[mid:])
    else:
        return findBin(n, bins[:mid])

with open('path/to/file') as infile:
    for line in infile:
        n = float(line.strip())
        counts[findBin(n, ranges)] += 1

for low,high in sorted(counts):
    print("There are", counts[(low,high)], "many numbers between", low, "and", high)

希望这有帮助

Answer 6

您可以使用以下代码

infile = open('text1.txt','r')
text = infile.read().split('\n')
infile.close()
#text.remove('')

#calculate the numbers of each range
numbers = [0,0,0,0,0]
for i in text:
    temp = float(i);
    temp = (temp-2.5)*100
    temp = int(temp)/20
    numbers[temp] =numbers[temp] + 1  

#display the numbers of each range
print 'Number of elements'
print '2.5-2.7: '+ str(numbers[0])
print '2.7-2.9: '+ str(numbers[1]) 
print '2.9-3.1: '+ str(numbers[2])
print '3.1-3.3: '+ str(numbers[3])
print '3.3-3.5: '+ str(numbers[4])

如何找到数字范围

6 个答案: