我正在用 panda 编写脚本,但我无法提取我想要的正确输出.这是问题:
I am writing scripts in panda but i could not able to extract correct output that i want. here it is problem:
我可以从 CSV 文件中读取这些数据.在这里你可以找到表结构
i can read this data from CSV file. Here you can find table structure
http://postimg.org/image/ie0od7ejr/
我想要上表数据的这个输出
I want this output from above table data
Month Demo1 Demo 2
June 2013 3 1
July 2013 2 2
在 Demo1 和 Demo2 列中,我想计算以 u 开头的常规条目和条目.六月共有 3 个常规条目,其中 1 个条目以 u 开头.
in Demo1 and Demo2 column i want to count regular entry and entry which starts with u. for June there are total 3 regular entry while 1 entry starts with u.
到目前为止,我已经编写了这段代码.
so far i have written this code.
import sqlite3
from pylab import *
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates
import datetime as dt
conn = sqlite3.connect('Demo2.sqlite')
df = pd.read_sql("SELECT * FROM Data", conn)
df['DateTime'] = df['DATE'].apply(lambda x: dt.date.fromtimestamp(x))
df1 = df.set_index('DateTime', drop=False)
感谢您的帮助.最终结果将是条形图.我可以从上面提到的输出中绘制图表.
Thanks advace for help. End result would be bar graph. I can draw graph from output that i mention above.
对于resample
,可以这样定义两个聚合函数:
For resample
, you can define two aggregation functions like this:
def countU(x):
return sum(i[0] == 'u' for i in x)
def countNotU(x):
return sum(i[0] != 'u' for i in x)
print df.resample('M', how=[countU, countNotU])
或者,考虑 groupby
.
这篇关于Python Panda TIME 系列重新采样的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持html5模板网!