目标是返回字符串中出现的3的倍数。我只需要担心3,6,9的倍数,并使用字典。例如,0939639将返回9,因为它出现了3次,而其他3的倍数出现的次数少于9。
下面是我的代码:
def count_threes(n):
# Turning the number into a string
x = len(str(n))
# Setting the keys(multiple of threes) to 0 as the default count
dict = {3: 0,
6: 0,
9: 0}
# L
我正在编写一本基于文本的RPG和一本名为“让您的Python文本冒险”的书,我遇到了一些问题。目前,游戏地图存储在字典中。但是,当我试图运行该程序时,我将在我的KeyError: ' '文件中获得一个world.py。
我已经尝试确定KeyError可能来自哪里,因为我知道这通常与字典IIRC中的条目有关。我无法找到这个错误的来源,通过查看错误所引用的代码行,我看不到KeyError可能来自何处。
for y, dsl_row in enumerate(dsl_lines):
row = []
dsl_cells = dsl_r
所以我正在开发一个程序,它过去可以很好地保存/加载,并且我有所有正确的导入等等,但是最近,当我尝试加载时,我收到了这个错误消息(在保存过程中也有类似的错误消息):
Traceback (most recent call last):
File "C:\Users\Adam\AppData\Local\Programs\Python\Python36\lib\shelve.py", line 111, in __getitem__
value = self.cache[key]
KeyError: 'flipvariables'
During ha
def user_info(request, template_name='social/retrieve_user_data.html', username=None):
user = User.objects.get(username=username)
if request.method == 'POST':
form = UserInfoForm(request.POST, user)
print(form.is_valid())
if form.is_valid():
我正在创建一个不和谐的机器人。需要先将命令参数发送到字典,然后再发送到json。首先作为键,然后作为数组。 但!它只发送第二个和第三个参数。 @client.command(pass_context=True)
async def next(ctx,name,time):
author = ctx.message.author
#role authentification
if "561139623250755585" in [y.id for y in author.roles]:
rere = time + '
while stack.isEmpty() != 1:
fin = stack.pop()
print fin - output is (1,1)
k = final.get(fin)
return k
def directionToVector(direction, speed = 1.0):
dx, dy = Actions._directions[direction]
return (dx * speed, dy * speed)
directionToVecto
如何返回只有key +value回文的字典。(赋值) def isPalindrome(dict1):
newDict = dict1.copy().items()
for key,value in newDict:
result = key + value
for index in range(len(result)):
if result[index] != result[len(result) - 1 - index]:
del dict1[key]
return dict1
dictionar
import numpy as np
#Collect the compound values for each news source
score_table = df.pivot_table(index='User', values="Compound", aggfunc = np.mean)
score_table
from collections import Counter
import pandas as pd
a = dict(Counter(HT_positive))
t = list(a.items())
compound = s
def load_data(label_dict):
x = []
y = []
x_keys = ['FEAT_1','FEAT_2','FEAT_3','FEAT_4','FEAT_5','FEAT_6']
y_keys = ['LABEL'];
for item in label_dict:
x.append([float(item[k]) for k in x_keys])
y.append([float
我正在尝试解析一个JSON文件,但得到以下错误:
Traceback (most recent call last):
File "data2spreadsheet.py", line 151, in <module>
parse(json.loads(line))
File "data2spreadsheet.py", line 119, in parse
tw.parse(tweet)
File "data2spreadsheet.py", line 78, in parse
self.url
我使用wang2vec ()预训练了一个单词嵌入,并通过gensim将其加载到python中。当我试图得到一些单词的向量时,我显然得到了:
KeyError: "word 'kjklk' not in vocabulary"
因此,我考虑在词汇表中添加一个项来映射oov (Oov)单词,比方说<OOV>。由于词汇表是Dict格式的,所以我只需添加项{"<OOV>":0}。
但是,我搜索了一项词汇
model = gensim.models.KeyedVectors.load_word2vec_format(w2v_ext,
CORPUS = [
'this is the first document',
'this is the second document',
'and this is the third document',
'is this the first document ?'
]
doc = CORPUS
dic = {}
for sentence in doc:
k = list(sentence.split())
for term in k:
我正面临着从字典中读取实例的问题。
我定义了一个字典,并按一定的顺序(标题、作者)添加条目。但我被要求以相反的形式检查它们(作者,标题),以检查它们的可用性。为了说明这一点,我举了一个图书实例:
class Book():
def __init__(self, title, author):
这是由一本书的字典组成的库类:
class Library:
def __init__(self, name):
self.name = name
#fill in here to define the books attribute/instance va
所以我的字典里有一个代码:
def get_rooms_for(dict1, num):
try:
for x in dict1:
if x == num:
print(dict1[x])
except KeyError:
print (num,"is not available.")
我的测试是
get_rooms_for({'CS101':3004, 'CS102':4501, 'CS103':6755,