应用函数创建多列作为参数的字符串

Question

我有这样的数据帧：

     name .  size . type    .  av_size_type
0    John .   23  . Qapra'  .            22
1     Dan .   21  . nuk'neH .            12
2  Monica .   12  . kahless .            15

我想创建一个带有句子的新列，如下所示：

    name .  size . type    .  av_size_type  .   sentence
0    John .   23 . Qapra'  .            22  .   "John has size 23, above the average of Qapra' type (22)"
1     Dan .   21 . nuk'neH .            12  .   "Dan has size 21, above the average of nuk'neH type (21)"
2  Monica .   12 . kahless .            15  .   "Monica has size 12l, above the average of kahless type (12)

它会是这样的：

def func(x):
    string="{0} has size {1}, above the average of {2} type ({3})".format(x[0],x[1],x[2],x[3])
    return string

df['sentence']=df[['name','size','type','av_size_type']].apply(func)

但是，显然这种类型的synthax不起作用。

有人会对此提出建议吗？

Answer 1

使用列表推导作为快速替代，因为您被迫迭代：

string = "{0} has size {1}, above the average of {2} type ({3})"
df['sentence'] = [string.format(*r) for r in df.values.tolist()]

df

     name  size     type  av_size_type  \
0    John    23   Qapra'            22   
1     Dan    21  nuk'neH            12   
2  Monica    12  kahless            15   

                                            sentence  
0  John has size 23, above the average of Qapra' ...  
1  Dan has size 21, above the average of nuk'neH ...  
2  Monica has size 12, above the average of kahle... 

Answer 2

您可以使用apply直接构建句子。

df['sentence'] = (
    df.apply(lambda x: "{} has size {}, above the average of {} type ({})"
                       .format(*x), axis=1)
)

如果要明确引用列，可以执行以下操作：

df['sentence'] = (
    df.apply(lambda x: "{} has size {}, above the average of {} type ({})"
                       .format(x.name, x.size, x.type, x.av_size_type), axis=1)
)

Answer 3

使用splat并打开包装

string = lambda x: "{} has size {}, above the average of {} type ({})".format(*x)

df.assign(sentence=df.apply(string, 1))

     name  size     type  av_size_type                                           sentence
0    John    23   Qapra'            22  John has size 23, above the average of Qapra' ...
1     Dan    21  nuk'neH            12  Dan has size 21, above the average of nuk'neH ...
2  Monica    12  kahless            15  Monica has size 12, above the average of kahle...

---

如果需要，可以使用字典解包

string = lambda x: "{name} has size {size}, above the average of {type} type ({av_size_type})".format(**x)

df.assign(sentence=df.apply(string, 1))

     name  size     type  av_size_type                                           sentence
0    John    23   Qapra'            22  John has size 23, above the average of Qapra' ...
1     Dan    21  nuk'neH            12  Dan has size 21, above the average of nuk'neH ...
2  Monica    12  kahless            15  Monica has size 12, above the average of kahle...