蟒蛇 - 熊猫:每
问题描述:
行
我有一个数据帧计数相同的值,如下:蟒蛇 - 熊猫:每
investing.com ft bloomberg
19 API Weekly Distillates Stocks NaN NaN
20 API Weekly Gasoline Stock NaN NaN
21 NaN Advance Goods Trade Balance Advance Goods Trade Balance
22 NaN NaN Advance Retail Sales
23 All Car Sales NaN NaN
24 All Truck Sales NaN NaN
25 Average Hourly Earnings MoM Average Hourly Earnings MoM Average Hourly Earnings MoM
26 NaN NaN Average Hourly Earnings YoY
我愿与所有未楠值的计数添加一列。
我想:
df['count of not NaN'] = df.apply(lambda x:(x[['investing.com','ft','bloomberg']] != 'NaN').count(), axis=1)
,但没有奏效。任何人都知道为什么/可以帮助我使用正确的配方? (我知道这个问题的一些口味已经发布,但他们不能真正帮助我获得成功的结果......) 谢谢!
答
计数方法正是这样做的。使用axis = 1添加一列。
df.count(axis=1)
是的..对不起,愚蠢的问题..我正要删除它,因为我刚才找到答案..反正。 –