Python fillna_python rfind函数

假设我有这张桌子

Type | Killed | Survived

Dog 5 2

Dog 3 4

Cat 1 7

Dog nan 3

cow nan 2

[Type] = Dog缺少Killed值之一.

我想在[类型] =狗的[Killed]中归咎于平均值.

我的代码如下：

>搜索平均值

df [df [‘Type’] ==’Dog’].mean().round()

这将给我平均值(约2.25)

>估算均值(这是问题开始的地方)

df.loc [(df [‘Type’] ==’Dog’)& (df [‘Killed’])].fillna(2.25,inplace = True)

代码运行,但值不是估算,NaN值仍然存在.

我的问题是,我如何根据[Type] = Dog来估算[Killed]中的均值.

最佳答案 对我来说工作：

df.ix[df[‘Type’] == ‘Dog’, ‘Killed’] = df.ix[df[‘Type’] == ‘Dog’, ‘Killed’].fillna(2.25)

print (df)

Type Killed Survived

0 Dog 5.00 2

1 Dog 3.00 4

2 Cat 1.00 7

3 Dog 2.25 3

4 cow NaN 2

如果系列需要fillna – 因为2列被杀和幸存：

m = df[df[‘Type’] == ‘Dog’].mean().round()

print (m)

Killed 4.0

Survived 3.0

dtype: float64

df.ix[df[‘Type’] == ‘Dog’] = df.ix[df[‘Type’] == ‘Dog’].fillna(m)

print (df)

Type Killed Survived

0 Dog 5.0 2

1 Dog 3.0 4

2 Cat 1.0 7

3 Dog 4.0 3

4 cow NaN 2

如果需要fillna只在Killed列中：

#if dont need rounding, omit it

m = round(df.ix[df[‘Type’] == ‘Dog’, ‘Killed’].mean())

print (m)

4

df.ix[df[‘Type’] == ‘Dog’, ‘Killed’] = df.ix[df[‘Type’] == ‘Dog’, ‘Killed’].fillna(m)

print (df)

Type Killed Survived

0 Dog 5.0 2

1 Dog 3.0 8

2 Cat 1.0 7

3 Dog 4.0 3

4 cow NaN 2

您可以重用以下代码：

filtered = df.ix[df[‘Type’] == ‘Dog’, ‘Killed’]

print (filtered)

0 5.0

1 3.0

3 NaN

Name: Killed, dtype: float64

df.ix[df[‘Type’] == ‘Dog’, ‘Killed’] = filtered.fillna(filtered.mean())

print (df)

Type Killed Survived

0 Dog 5.0 2

1 Dog 3.0 8

2 Cat 1.0 7

3 Dog 4.0 3

4 cow NaN 2