Matplotlib-显示唯一值频率的条形图

2024-10-05 14:29:33 发布

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我有一个数组,比如:

[1,3,5,2,12,12,3,6,…]等,在代表蛾类的1-12之间变化。

现在,我可以用

plt.hist(months,range(1, 13))
plt.title("Job Request per month")
plt.xlabel("Value")
plt.ylabel("Frequency")

plt.show()

现在,它显示直方图上唯一值的计数。然而,我找不到一个方法来与条形图。

完整代码:

from pymongo import MongoClient
import matplotlib.pyplot as plt

def main():

client = MongoClient('mongodb://a12345:a12345@127.0.0.1:27017/')
db = client['newDatabase']
collection = db['jobs']
cursor = collection.find({}, {'_constructed': 1})

months = []
for document in cursor:
    if '_constructed' in document: 
        months.append(document['_constructed'].month)

print(months);
plt.bar(months, range(1, 13))
plt.title("Job Request per month")
plt.xlabel("Value")
plt.ylabel("Frequency")

plt.show()

if __name__ == "__main__":
    main()

Tags: titlevaluemainrequestjobrangepltdocument
2条回答

试试这个:

#!python
# -*- coding: utf-8 -*-#
#
# Imports
import matplotlib.pyplot as plt
import numpy as np
import collections

data = """
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
"""
months = data.split()

data = [6, 6, 6, 6, 6, 6, 6, 6, 1, 1, 6, 1, 6, 2, 1, 1, 6, 1, 6, 2, 1, 6, 6, 2, 1, 2, 1, 6, 2, 1, 1, 6, 1, 6, 3, 3, 3, 3, 3, 1, 2, 1, 3, 3, 3, 3, 3, 1, 2, 1, 3, 3, 3, 3, 3, 1, 2, 1, 6, 2, 1, 10,
        6, 12, 9, 3, 3, 3, 3, 3, 1, 2, 1, 7, 8, 9, 10, 10, 11, 12, 11, 3, 3, 3, 3, 3, 1, 12, 12, 6, 2, 5, 10, 6, 12, 9, 3, 3, 3, 3, 5, 1, 2, 1, 7, 8, 9, 10, 10, 11, 12, 11, 5, 5, 3, 3, 3, 1, 12, 12]

c = collections.Counter(data)
c = sorted(c.items())
months_num = [i[0] for i in c]
month_names = [months[i[0]-1] for i in c]
freq = [i[1] for i in c]

print(c)
print(months)
print(freq)
f, ax = plt.subplots()


plt.bar(months_num, freq)
plt.title("Job Request per month")
plt.xlabel("Months")
plt.ylabel("Frequency")
ax.set_xticks(range(1, 13))
ax.set_xticklabels(months)

plt.show()

enter image description here

您可以使用np.bincount先计算months,然后绘制:

import matplotlib.pyplot as plt
import numpy as np

# create an example months array
months = np.random.randint(1, 13, size=30)

plt.bar(np.arange(1, 13), np.bincount(months, minlength=13)[1:])

enter image description here

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