Python 应用函数创建以多列作为参数的字符串

我有这样一个数据帧：

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

    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)

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)

我想用一个句子创建一个新列，如下所示：

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

    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)

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)

应该是这样的：

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

    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)

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)

然而，显然这种synthax不起作用

有人对此有什么建议吗？

使用splat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...

---

如果你愿意，你可以使用字典解包

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...

---

使用列表理解作为快速替代，因为您必须迭代：

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

---

你可以使用apply直接构建句子

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

如果要显式引用列，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)
)

---

你忘了在函数中返回字符串…不知道，在fn中尝试

返回字符串

（也许可以睡一会儿：）@umutto-oops，没错。已经修好了。这两种方法都不起作用耶，我刚刚意识到你也需要在列上应用，所以

.apply（func，axis=1）

应该可以。谢谢@Allen。这似乎是个好办法。有没有一种方法可以选择我想在格式中放置哪些列？我的df实际上有几十个列，我试图在这里简化它。谢谢@piRSquared！有没有一种方法可以选择我想在格式中放置哪些列？我的df实际上有几十列，我在这里试图简化它。您可以将行序列用作字典，并使用双splat进行解包。@OP，使用此方法，选择要输出的列是最简单的。只需按如下方式索引它们：

df[col1，col2，…].values.tolist（）