本文介绍了根据多种条件格式化 pandas 数据框中的单元格颜色的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试格式化数据框中特定列的单元格的颜色,但是我无法根据多种条件来做到这一点.
I am trying to format the color of a cell of an specific column in a data frame, but I can't manage to do it according to multiple conditions.
这是我的数据框(df):
This is my dataframe (df):
Name ID Cel Date
0 Diego b000000005 7878 2565-05-31 20:53:00
1 Luis b000000015 6464 2017-05-11 20:53:00
2 Vidal b000000002 1100 2017-05-08 20:53:00
3 John b000000011 4545 2017-06-06 20:53:00
4 Yusef b000000013 1717 2017-06-06 20:53:00
我希望日期"列中的值根据以下条件更改颜色:
I want the values in the "Date" column to change color according to the following conditions:
if date < datetime.now():
color = 'green'
elif date > datetime.now():
date = 'yellow'
elif date > (datetime.now() + timedelta(days=60)):
color = 'red'
这是我当前的代码:
def color(val):
if val < datetime.now():
color = 'green'
elif val > datetime.now():
color = 'yellow'
elif val > (datetime.now() + timedelta(days=60)):
color = 'red'
return 'background-color: %s' % color
df.style.apply(color, subset = ['Fecha'])
我遇到以下错误:
输出为:
Out[65]: <pandas.formats.style.Styler at 0x1e3ab8dec50>
任何帮助将不胜感激.
推荐答案
使用 applymap
:
from datetime import datetime, timedelta
import pandas as pd
name = ['Diego', 'Luis', 'Vidal', 'John', 'Yusef']
id = ['b000000005', 'b000000015', 'b000000002', 'b000000011', 'b000000013']
cel = [7878, 6464, 1100, 4545, 1717]
date = pd.to_datetime(['2017-05-31 20:53:00', '2017-05-11 20:53:00', '2017-05-08 20:53:00',
'2017-06-06 20:53:00', '2017-06-06 20:53:00'])
df = pd.DataFrame({'Name':name,'ID':id,'Cel':cel,'Date':date})
def color(val):
if val < datetime.now():
color = 'green'
elif val > datetime.now():
color = 'yellow'
elif val > (datetime.now() + timedelta(days=60)):
color = 'red'
return 'background-color: %s' % color
df.style.applymap(color, subset=['Date'])
Jupyter笔记本的屏幕截图.如果您改为print
输出,则只会得到对Styler
对象的引用:
Screenshot from Jupyter notebook. If you print
the output instead, you'll just get a reference to the Styler
object:
print(df.style.applymap(color, subset=['Date']))
<pandas.formats.style.Styler object at 0x116db43d0>
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