Python 3.x 如何根据条件删除熊猫中的行？

我有以下数据框

df = pd.DataFrame([['1','aa','ccc','rere','thth','my desc 1','','my feature2 1'], ['1','aa','fff','flfl','ipip','my desc 2','',''], ['1','aa','mmm','rprp','','','',''], ['2','aa','ccc','rprp','','','my feature1 1',''], ['2','aa','fff','bubu','thth','my desc 3','',''], ['2','aa','mmm','fafa','rtrt','my desc 4','',''], ['3','aa','ccc','blbl','thth','my desc 5','my feature1 2','my feature2 2'], ['3','aa','fff','arar','amam','my desc 6','',''], ['3','aa','mmm','acac','ryry','my desc 7','',''],['4','bb','coco','rere','','','','my feature2 3'], ['4','bb','inin','mimi','rere','my desc 8','',''], ['4','bb','itit','toto','enen','my desc 9','',''], ['4','bb','spsp','glgl','pepe','my desc 10','',''], ['5','bb','coco','baba','mpmp','my desc 11','my feature1 3',''], ['5','bb','inin','rere','','','',''],['5','bb','itit','toto','hrhr','my desc 12','',''], ['5','bb','spsp','glgl','lolo','my desc 13','','']], columns=['foo', 'bar','name_input','value_input','bulb','desc','feature1', 'feature2'])

现在，我需要删除行以获得下面的输出

df = pd.DataFrame([['1','aa','ccc','rere','thth','my desc 1','','my feature2 1'], ['2','aa','ccc','rprp','','my desc 3','my feature1 1',''], ['3','aa','ccc','blbl','thth','my desc 5','my feature1 2','my feature2 2'], ['4','bb','coco','rere','','my desc 8','','my feature2 3'], ['5','bb','coco','baba','mpmp','my desc 11','my feature1 3','']], columns=['foo', 'bar','name_input','value_input','bulb','desc','feature1', 'feature2'])

我试过下面的方法。它们似乎都不起作用

df= df.dropna(subset=['feature1', 'feature2'])
df.dropna(thresh=5, axis=0, inplace=True)
df= df[df.feature2.notnull()]
df= df[pd.notnull(df[['feature1', 'feature2']])]

非常感谢您的帮助

astype（bool）

在布尔上下文中，空字符串的计算结果为

False

。使用

filter

仅获取以

feature

开头的列。然后使用

astype（bool）

，后跟

any（axis=1）

为了匹配您的结果，我们可以倒填

desc

列

feat = df.filter(regex='feat').astype(bool).any(1)
desc = df.desc.where(df.desc.astype(bool)).bfill()
df.assign(desc=desc)[feat]

   foo bar name_input value_input  bulb        desc       feature1       feature2
0    1  aa        ccc        rere  thth   my desc 1                 my feature2 1
3    2  aa        ccc        rprp         my desc 3  my feature1 1               
6    3  aa        ccc        blbl  thth   my desc 5  my feature1 2  my feature2 2
9    4  bb       coco        rere         my desc 8                 my feature2 3
13   5  bb       coco        baba  mpmp  my desc 11  my feature1 3               

---

另一种方法是将空白字符串更改为true

NaN

值，然后将

how

参数传递给

dropna

并使用

all

作为值

import numpy as np
df.replace('',np.nan).dropna(subset=['feature1','feature2'],how='all').fillna('')


   foo bar name_input value_input  bulb        desc       feature1  feature2
0    1  aa        ccc        rere  thth   my desc 1                 my feature2 1
3    2  aa        ccc        rprp                    my feature1 1   
6    3  aa        ccc        blbl  thth   my desc 5  my feature1 2  my feature2 2
9    4  bb       coco        rere                                   my feature2 3 
13   5  bb       coco        baba  mpmp  my desc 11  my feature1 3  

---

非常圆滑，不知道字符串求值为False，谢谢你的解释。谢谢。但是，“desc”列应该包含所有值，包括“my desc 3”和“my desc 8”。我们是如何得到它的？你需要解释为什么“我的描述3”是这样的。对于单个的foo，描述值是强制性的。和特征1、特征2取决于灯泡。如果灯泡为空，则其中一个功能将为空，但desc将始终显示值。需要每个foo的第一个可用描述

import numpy as np
df.replace('',np.nan).dropna(subset=['feature1','feature2'],how='all').fillna('')


   foo bar name_input value_input  bulb        desc       feature1  feature2
0    1  aa        ccc        rere  thth   my desc 1                 my feature2 1
3    2  aa        ccc        rprp                    my feature1 1   
6    3  aa        ccc        blbl  thth   my desc 5  my feature1 2  my feature2 2
9    4  bb       coco        rere                                   my feature2 3 
13   5  bb       coco        baba  mpmp  my desc 11  my feature1 3