熊猫得到错误的结果时<；添加然后比较>；数字，python

当我需要删除数据帧中“所有概率值之和（前10位浮点数）大于1”的行时，pandas给了我错误的结果

我的代码：

# drop wrong probability row
data.at[data[data.p1 + data.p2 + data.p3 > 1.001].index, 'h1'] = 'dropped by pandas'

结果是：

re______________ | p1________   | p2________ | p3________ | sump

correct result_____ | 0.743088844   | 0.24208727    | 0.014823886   | 1 << correct
correct result_____ | 0.647239626   | 0.346835025   | 0.00592535    | 1 << correct
correct result_____ | 0.65043824    | 0.34372226    | 0.0058395 | 1 << correct
correct result_____ | 0.75111312    | 0.221604341   | 0.027282539   | 1  << correct
dropped by pandas   | 0.670277591   | 0.324265434   | 0.005456975   | 1  << wrong
dropped by pandas   | 0.672221755   | 0.322438072   | 0.005340173   | 1  << wrong
dropped by pandas   | 0.670053332   | 0.324742569   | 0.005204099   | 1  << wrong
dropped by pandas   | 0.667690433   | 0.327033634   | 0.005275932   | 1  << wrong
dropped by pandas   | 0.237037933   | 0.823248091   | 0.05335034    | 1.113636364  << correct
dropped by pandas   | 0.242720919   | 0.818282268   | 0.052633177   | 1.113636364  << correct

re uuuuuuuuuuuuuuuuuuu124; p1 uuuuuuuuuuuu124; p2 uuuuuuuuuu124; p3 uuuuuuuu124;集水坑
正确的结果uuuuuuuuuuuuuuuuu|0.743088844 | 0.24208727 | 0.014823886 | 1您将所有结果相加，然后将它们与一进行比较，而不是将它们除以3或与3进行比较。
只需将
data.at[data[data.p1+data.p2+data.p3>1.001]。索引“h1']”改为
data.at[data[data.p1+data.p2+data.p3>3]。索引“h1']=”被熊猫丢弃“
或
数据。索引[data[data.p1+data.p2+data.p3/3>1]。索引“h1'=”被熊猫丢弃“
。此外，您不需要将它们与1.001进行比较，您可以将它们与1进行比较，因为>函数大于大于等于>=。。。有时它会工作，但有时它不会。。。请原谅我的打字错误=谢谢你的回答：）实际上p1，p2，p3的和已经等于1了。我将其与1.001进行比较的原因是，正如你所说，当我将其与1进行比较时，pandas在我的数据帧中删除了更多行。另一方面，当我们找出问题所在时，我愿意将它改为1。