方法一:Linq ChannelList就是一个List类型的数据,IsOpen 是其元素的属性 channelCount = (from channel in DevicesManager.Instance.CurrentDevice.ChannelList where channel.IsOpen group channel by channel.ChannelID).Count(); 方法二:泛型委托Predicate<T> public delegate bool Predicate&
删除df中任意字段等于'null'字符串的行: df=df.astype(str)#把df所有元素转为str类型 df=df[df['A'].isin(['null','NULL'])] #找出df的'A'列值为'null'或'NULL'(注意此处的null是字符串,不是空值) df=df[~df['A'].isin(['null','NULL'])] #过滤掉A列为'null'或'NULL'的行,~表示取反 去掉任意一列为'null'值的行,目前只能想到用循环: for col in list
这样用,只会替换匹配到的第一个子串 str = 'I hava a pen ,I hava an apple,apple pen, pen apple' str = str.replace('apple', 'pear') //str='I hava a pen ,I hava an pear,apple pen, pen apple' 加上全局标识(g),替换全部子串 str = 'I hava a pen ,I hava an apple,apple pen, pen apple' str