一位 friend 向我发送了这个虚拟代码,以便做一个简单的密码:
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']
所以他尝试迭代这个单词来对其进行编码:
word1="helloworld"
for char in word1:
h=code.loc[code['Key'] == char, 'Value'][0]
语法看起来相同,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。也许我错过了一些东西,它来自 Pandas 。我正在运行版本'0.20.3'
编辑:当我发布时,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,我完全回避了我的帖子的要点。我想了解为什么它在这个简单的情况下有效,而不是在迭代期间有效。
最佳答案
只需删除 h 属性末尾的 [0]:
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)
我添加了两个打印来确保代码确实做了他应该做的事情,我得到了以下结果:
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
关于python - KeyError 0 在 pandas 中选择带有字符串的位置,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46288216/