python - KeyError 0 loc 选择与 Pandas 中的字符串

Question

一个 friend 发给我这个伪代码，为了做一个简单的密码:

chaine1="abcdefghijklmnopqrstuvwxyz"
chaine2="abcdefghijklmnopqrstuvwxyz"
list1=list(chaine1)
list2=list(chaine2)
rnd.shuffle(list2)
code=pd.DataFrame({"Key": list1, "Value" : list2})

他设法恢复了这样一封信:

a = code.loc[code['Key'] == 'a', 'Value']

所以他试图遍历这个词来对其进行编码:

word1="helloworld"
for char in word1:
    h=code.loc[code['Key'] == char, 'Value'][0]

语法看起来一样，但失败了:

KeyError                                  Traceback (most recent call last)
<ipython-input-88-4e2a59e0978e> in <module>()
      1 word1="helloworld"
      2 for char in word1:
----> 3     h=code.loc[code['Key'] == char, 'Value'][0]

~/Envs/test_bapt/lib/python3.5/site-packages/pandas/core/series.py in __getitem__(self, key)
    599         key = com._apply_if_callable(key, self)
    600         try:
--> 601             result = self.index.get_value(self, key)
    602 
    603             if not is_scalar(result):

~/Envs/test_bapt/lib/python3.5/site-packages/pandas/core/indexes/base.py in get_value(self, series, key)
   2475         try:
   2476             return self._engine.get_value(s, k,
-> 2477                                           tz=getattr(series.dtype, 'tz', None))
   2478         except KeyError as e1:
   2479             if len(self) > 0 and self.inferred_type in ['integer', 'boolean']:

pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_value (pandas/_libs/index.c:4404)()

pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_value (pandas/_libs/index.c:4087)()

pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc (pandas/_libs/index.c:5126)()

pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.Int64HashTable.get_item (pandas/_libs/hashtable.c:14031)()

pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.Int64HashTable.get_item (pandas/_libs/hashtable.c:13975)()

KeyError: 0

我发现 .values 丢失了:h=code.loc[code['Key'] == char, 'Value'][0]。但是有人知道为什么第一行有效吗？我认为迭代一个字符串仍然会返回一个字符串。也许我遗漏了一些东西，它来自 Pandas 。我正在运行版本 '0.20.3'

编辑: 当我发帖时，我忘记了 a 定义中的 [0] 它应该是:

a = code.loc[code['Key'] == 'a', 'Value'][0]

很抱歉，我完全避开了我帖子的重点。我想了解为什么它在这个简单的案例中有效，而不是在迭代期间。

Answer 1

只需去掉h属性末尾的[0]即可:

import random as rnd
import pandas as pd

chaine1="abcdefghijklmnopqrstuvwxyz"
chaine2="abcdefghijklmnopqrstuvwxyz"
list1=list(chaine1)
list2=list(chaine2)
rnd.shuffle(list2)
code=pd.DataFrame({"Key": list1, "Value" : list2})

word1="helloworld"
for char in word1:
    print(char)
    h=code.loc[code['Key'] == char, 'Value']
    print(h)

我添加了两个打印以确保代码确实按照他应该的方式执行，我得到以下结果:

h
7    e
Name: Value, dtype: object
e
4    m
Name: Value, dtype: object
l
11    t
Name: Value, dtype: object
l
11    t
Name: Value, dtype: object
o
14    j
Name: Value, dtype: object
w
22    d
Name: Value, dtype: object
o
14    j
Name: Value, dtype: object
r
17    i
Name: Value, dtype: object
l
11    t
Name: Value, dtype: object
d
3    f
Name: Value, dtype: object