如何垃圾桶系列浮点值到柱状图在Python?
-
19-09-2019 - |
题
我在浮子(总是小于0)设定的值。我想斌到柱状图, I,E。在直方图中的每个条包含的值的范围[0,0.150)
我的数据看起来像这样:
0.000
0.005
0.124
0.000
0.004
0.000
0.111
0.112
蒙山我的代码下面我期望得到的结果,看起来像
[0, 0.005) 5
[0.005, 0.011) 0
...etc..
我试图做做这种合并与我的这个代码。 但它似乎并没有工作。什么是应该做的正确方法?
#! /usr/bin/env python
import fileinput, math
log2 = math.log(2)
def getBin(x):
return int(math.log(x+1)/log2)
diffCounts = [0] * 5
for line in fileinput.input():
words = line.split()
diff = float(words[0]) * 1000;
diffCounts[ str(getBin(diff)) ] += 1
maxdiff = [i for i, c in enumerate(diffCounts) if c > 0][-1]
print maxdiff
maxBin = max(maxdiff)
for i in range(maxBin+1):
lo = 2**i - 1
hi = 2**(i+1) - 1
binStr = '[' + str(lo) + ',' + str(hi) + ')'
print binStr + '\t' + '\t'.join(map(str, (diffCounts[i])))
〜
解决方案
当可能,不要重新发明轮子。 NumPy的有你需要的一切:
#!/usr/bin/env python
import numpy as np
a = np.fromfile(open('file', 'r'), sep='\n')
# [ 0. 0.005 0.124 0. 0.004 0. 0.111 0.112]
# You can set arbitrary bin edges:
bins = [0, 0.150]
hist, bin_edges = np.histogram(a, bins=bins)
# hist: [8]
# bin_edges: [ 0. 0.15]
# Or, if bin is an integer, you can set the number of bins:
bins = 4
hist, bin_edges = np.histogram(a, bins=bins)
# hist: [5 0 0 3]
# bin_edges: [ 0. 0.031 0.062 0.093 0.124]
其他提示
from pylab import *
data = []
inf = open('pulse_data.txt')
for line in inf:
data.append(float(line))
inf.close()
#binning
B = 50
minv = min(data)
maxv = max(data)
bincounts = []
for i in range(B+1):
bincounts.append(0)
for d in data:
b = int((d - minv) / (maxv - minv) * B)
bincounts[b] += 1
# plot histogram
plot(bincounts,'o')
show()
第一个错误是:
Traceback (most recent call last):
File "C:\foo\foo.py", line 17, in <module>
diffCounts[ str(getBin(diff)) ] += 1
TypeError: list indices must be integers
为什么你一个int需要一个STR时转换为STR?解决这个问题,那么我们得到:
Traceback (most recent call last):
File "C:\foo\foo.py", line 17, in <module>
diffCounts[ getBin(diff) ] += 1
IndexError: list index out of range
因为你只取得5桶。我不明白你的桶装方案,但让我们把它50桶,看看会发生什么:
6
Traceback (most recent call last):
File "C:\foo\foo.py", line 21, in <module>
maxBin = max(maxdiff)
TypeError: 'int' object is not iterable
maxdiff
是单值超出你的整数列表中,那么,什么是max
在这里做什么?删除它,现在我们得到:
6
Traceback (most recent call last):
File "C:\foo\foo.py", line 28, in <module>
print binStr + '\t' + '\t'.join(map(str, (diffCounts[i])))
TypeError: argument 2 to map() must support iteration
果然,您使用的一个单一的值作为第二个参数,以map
。让我们简化了最后两行从这样的:
binStr = '[' + str(lo) + ',' + str(hi) + ')'
print binStr + '\t' + '\t'.join(map(str, (diffCounts[i])))
这样:
print "[%f, %f)\t%r" % (lo, hi, diffCounts[i])
现在它打印:
6
[0.000000, 1.000000) 3
[1.000000, 3.000000) 0
[3.000000, 7.000000) 2
[7.000000, 15.000000) 0
[15.000000, 31.000000) 0
[31.000000, 63.000000) 0
[63.000000, 127.000000) 3
我不知道其他人在这里做什么,因为我真的不明白你希望使用了瓢泼大雨。这似乎涉及二元权力,但并不决策意识给我...
不隶属于 StackOverflow