date rtn 0 2016-01-01 0.101536 1 2016-01-02 0.055109 2 2016-01-03 -0.163625 3 2016-01-04 0.086756 4 2016-01-05 0.035907 5 2016-01-06 0.047619 6 2016-01-07 0.018717 7 2016-01-08 -0.036290
这样子的 df
大佬帮帮忙，实在搞不出来了
推荐学习书目
› Learn Python the Hard Way
Python Sites
› PyPI - Python Package Index
› http://diveintopython.org/toc/index.html
› Pocoo
值得关注的项目
› PyPy
› Celery
› Jinja2
› Read the Docs
› gevent
› pyenv
› virtualenv
› Stackless Python
› Beautiful Soup
› 结巴中文分词
› Green Unicorn
› Sentry
› Shovel
› Pyflakes
› pytest
Python 编程
› pep8 Checker
Styles
› PEP 8
› Google Python Style Guide
› Code Style from The Hitchhiker's Guide
This topic created in 2812 days ago, the information mentioned may be changed or developed.
6 replies • 2018-09-29 15:32:59 +08:00
 |
1
neosfung Sep 28, 2018
df['rtn'] = df['rtn'].apply(lambda x: 1 if x>0 else x)
|
 |
2
jswangjieda Sep 28, 2018
df.loc[df['rtn']>0,'rtn']=1
|
 |
3
ddzzhen Sep 28, 2018 via Android
直接改为 true，false 矩阵
|
 |
4
lieh222 Sep 29, 2018
x = x**0
|
 |
6
LaFayette Sep 29, 2018
df['rtn'] = np.where(df['rtn']>0, 1, df['rtn'])
|
Select Language