如何在fstrings中使用.loc?

时间:2018-10-04 09:50:45

标签: python string pandas indexing f-string

我有一个这样的数据框:

import pandas as pd

df = pd.DataFrame({'col1': ['abc', 'def', 'tre'],
                   'col2': ['foo', 'bar', 'stuff']})

  col1   col2
0  abc    foo
1  def    bar
2  tre  stuff

和这样的字典:

d = {'col1': [0, 2], 'col2': [1]}

字典包含列名和值索引,这些值将从数据框中提取以生成如下字符串:

abc (0, col1)

因此,每个字符串都以元素本身开头,并在括号中显示索引和列名。

我尝试了以下列表理解:

l = [f"{df.loc[{indi}, {ci}]} ({indi}, {ci})"
     for ci, vali in d.items()
     for indi in vali]

产生

['  col1\n0  abc (0, col1)',
 '  col1\n2  tre (2, col1)',
 '  col2\n1  bar (1, col2)']

因此,几乎可以,只需避免使用col1\n0部分。

如果我尝试

f"{df.loc[0, 'col1']} is great"

我明白了

'abc is great'

根据需要

x = 0
f"{df.loc[{x}, 'col1']} is great"

我明白了

'0    abc\nName: col1, dtype: object is great'

如何解决?

2 个答案:

答案 0 :(得分:2)

import pandas as pd

df = pd.DataFrame({'col1': ['abc', 'def', 'tre'],
                   'col2': ['foo', 'bar', 'stuff']})

d = {'col1': [0, 2], 'col2': [1]}
x = 0
[f"{df.loc[x, 'col1']} is great"
     for ci, vali in d.items()
     for indi in vali]

为您提供:

['abc is great', 'abc is great', 'abc is great']

这是您要找的吗?

您也可以遍历x范围

[f"{df.loc[i, 'col1']} is great"
 for ci, vali in d.items()
 for indi in vali
 for i in range(2)]

#output
['abc is great',
 'def is great',
 'abc is great',
 'def is great',
 'abc is great',
 'def is great']

答案 1 :(得分:1)

您所看到的是\n祖先返回的pd.Series对象的字符串表示形式和丑陋的换行符loc字符。

您应该使用pd.DataFrame.at返回标量,并且请注意,这里不需要为索引标签嵌套{}

L = [f'{df.at[indi, ci]} ({indi}, {ci})' \
     for ci, vali in d.items() \
     for indi in vali]

print(L)

['abc (0, col1)', 'tre (2, col1)', 'bar (1, col2)']