本文介绍了查找最接近输入值的行的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在努力解决这个问题.我有一个看起来像这样的熊猫数组

I am struggling with this problem. I have an pandas array which looks like this

               delta_n    column_1   ...
0                10      10        ...
1                20       0
2                30       0

现在我有一个数字,假设我搜索 delta_n=20.5 并且我想选择最接近 delta_n 数字的行.

Now i have a number, lets say i search for delta_n=20.5 and I want to select the row closest to the number of delta_n.

我的输出应该是:

1                20       0

我用 df.loc[20.5] 尝试过,但由于它不在 pd 数据帧中,所以它不起作用.

I tried it with df.loc[20.5] but as it is not in the pd dataframe it doesn't work.

谢谢,R

推荐答案

通过 sub,通过abs,通过idxmin 和最后选择 loc:

idx = df['delta_n'].sub(delta_n).abs().idxmin()

#added double [[]] for one row DataFrame
df1 = df.loc[[idx]]
print (df1)
   delta_n  column_1
1       20         0

#output Series with one []
s = df.loc[idx]
print (s)
delta_n     20
column_1     0
Name: 1, dtype: int64

详情:

print (df['delta_n'].sub(delta_n))
0   -10.5
1    -0.5
2     9.5
Name: delta_n, dtype: float64

print (df['delta_n'].sub(delta_n).abs())
0    10.5
1     0.5
2     9.5
Name: delta_n, dtype: float64

print (df['delta_n'].sub(delta_n).abs().idxmin())
1

numpy.argmin 的另一种 numpy 职位解决方案 并通过 iloc:

pos = df['delta_n'].sub(delta_n).abs().values.argmin()
print (pos)
1

df1 = df.loc[[pos]]
print (df1)
   delta_n  column_1
1       20         0

s = df.loc[pos]
print (s)
delta_n     20
column_1     0
Name: 1, dtype: int64

这篇关于查找最接近输入值的行的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

05-24 03:44