标签: keyerror

当一个数据帧中的日期时间对象位于另一个数据帧的日期时间对象范围内时，尝试合并两个数据帧。

继续获取：KeyError: 'cannot use a single bool to index into setitem' 在我发布的第二块代码中。

gametaxidf.loc[arrivemask, 'relevant'] = 1

我假设它也会在下面的行中使用类似的命令发生。

这是给我带来麻烦的部分：

with open('/Users/benjaminprice/Desktop/TaxiCombined/Data/combinedtaxifiltered.csv', 'w') as csvfile: 
    fieldnames1 = ['index','pickup_datetime', 'dropoff_datetime', 'pickup_long', 'pickup_lat','dropoff_long','dropoff_lat','passenger_count','trip_distance','fare_amount','tip_amount','total_amount','stadium_code'] 
    writer = csv.DictWriter(csvfile, fieldnames=fieldnames1) 
    writer.writeheader()

for index, row in baseballdf.iterrows(): 
    gametimestart = row['Start.Time'] 
    gametimeend = row['End.Time'] 
    arrivemin = gametimestart - datetime.timedelta(minutes=120) 
    arrivemax = gametimeend - datetime.timedelta(minutes = 30) 
    departmin = gametimeend - datetime.timedelta(minutes = 60) 
    departmax = gametimeend + datetime.timedelta(minutes = 90)

    gametaxidf = combineddf[combineddf.DATE==row.DATE]
    gametaxidf['relevant']=0

    for index, row in gametaxidf.iterrows(): …

尝试运行包含以下代码的脚本以生成文本块时:

from textwrap import dedent

text = dedent("""\
   yada yada '1' ('2','3',4') 
   ('{0}', Null, '{1}',
   '{
      "Hello":"world",
    }', '1', '{2}');""").format("yada1","yada2","yada3")

我得到一致的错误KeyError '\n "Hello"
并追溯指向的线.format().

当我删除format一切都没问题,但我需要它动态输入参数.
(最初它位于循环内)

我有一个字典dct,我希望它的每个值都可以求和,只要它们在指定的列表中存在相应的键lst.

我到目前为止使用的代码是:

sum(dct[k] for k in lst)

在上面的生成器表达式中,我想处理KeyError以防在列表中找不到列表中的键.我似乎无法找到如何实现(语法明智的)或者是try- except的方式,也不是if- else这个生成器表达式内的方法.

如果在字典中找不到列表中的键,则它应该继续获取其他值.总和的最终结果不应受任何缺失键的影响.如果没有密钥存在,那么零应该是和的结果.

我从这里运行代码:

import plotly
import plotly.plotly as py
from plotly.tools import FigureFactory as FF

import numpy as np
import pandas as pd
print(plotly.__version__)
dataframe = pd.DataFrame(np.random.randn(100, 3),
                         columns=['Column A', 'Column B', 'Column C'])

fig = FF.create_scatterplotmatrix(dataframe, diag='histogram', index='Column A',
                                  colormap=['rgb(100, 150, 255)', '#F0963C', 'rgb(51, 255, 153)'],
                                  colormap_type='seq', height=800, width=800)
py.iplot(fig, filename = 'Custom Sequential Colormap')

我收到这个错误:

/Library/Frameworks/Python.framework/Versions/3.4/bin/python3.4 /Users/mona/PycharmProjects/PythonCodes/plotly_viz.py
1.12.4
This is the format of your plot grid:
[ (1,1) x1,y1 ]  [ (1,2) x2,y2 ]
[ (2,1) x3,y3 ] …

SQLAlchemy 很好地记录了如何将关联对象与back_populates.

但是，当从该文档复制并粘贴示例时，将子项添加到父项会抛出 a ，KeyError如以下代码所示。模型类 100% 从文档中复制：

from sqlalchemy import Column, ForeignKey, Integer, String
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship
from sqlalchemy.schema import MetaData

Base = declarative_base(metadata=MetaData())

class Association(Base):
    __tablename__ = 'association'
    left_id = Column(Integer, ForeignKey('left.id'), primary_key=True)
    right_id = Column(Integer, ForeignKey('right.id'), primary_key=True)
    extra_data = Column(String(50))
    child = relationship("Child", back_populates="parents")
    parent = relationship("Parent", back_populates="children")

class Parent(Base):
    __tablename__ = 'left'
    id = Column(Integer, primary_key=True)
    children = relationship("Association", back_populates="parent")

class Child(Base):
    __tablename__ = 'right'
    id = …

我KeyError在使用collections.defaultdictwith.format()方法时得到了

外壳执行

In [1]: from collections import defaultdict                                                                                                                                                                        

In [2]: foo = "Foo: {foo}\nBar: {bar}"                                                                                                                                                                             

In [3]: default = defaultdict(lambda: 0)                                                                                                                                                                           

In [4]: foo.format(**default)                                                                                                                                                                                      
---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-4-799cab129cf2> in <module>
----> 1 foo.format(**default)

KeyError: 'foo'

我不期待 aKeyError因为我使用的是defaultdict. 为什么会发生这种情况？

除此之外，我想foo用一些默认值填充变量，我该怎么做？

只是试图从一堆照片的 EXIF 数据中提取一些纬度/经度信息，但代码会抛出 a ，KeyError即使稍后（成功）使用该键来打印特定坐标。

有问题的字典是“ tags” -'GPS GPSLatitude'并且'GPS GPSLongitude'都是tags.keys(); 我已经三重检查了。

那么对于为什么会抛出关键错误有什么直觉tags['GPS GPSLatitude']吗tags['GPS GPSLongitude']？

import os
import exifread

output = dict()
output['name'] = []
output['lon'] = []
output['lat'] = []

for file in os.listdir(path):
    if file.endswith(".JPG"):
        full_path = path + file
        print (file) #check to ensure all files were found
        output['name'].append(file) #append photo name to dictionary
        f = open(full_path, 'rb') #open photo
        tags = exifread.process_file(f) #read exifdata …

我是Python的新手,我想用Python来复制一个常见的excel任务.如果已经回答了这样的问题,请告诉我.我一直无法找到它.我有以下pandas数据帧(数据):

Date    Stage   SubStage    Value
12/31/2015   1.00   a   0.896882891
1/1/2016     1.00   a   0.0458843
1/2/2016     1.00   a   0.126805588
1/3/2016     1.00   b   0.615824461
1/4/2016     1.00   b   0.245092069
1/5/2016     1.00   c   0.121936318
1/6/2016     1.00   c   0.170198128
1/7/2016     1.00   c   0.735872415
1/8/2016     1.00   c   0.542361912
1/4/2016     2.00   a   0.723769247
1/5/2016     2.00   a   0.305570257
1/6/2016     2.00   b   0.47461605
1/7/2016     2.00   b   0.173702623
1/8/2016     2.00   c   0.969260251
1/9/2016     2.00   c   0.017170798

在excel中,我可以使用数据透视表来生成以下内容:

使用'数据'的excel数据透视表

在python中执行以下操作似乎是合理的:

data.pivot(index='Date',columns = ['Stage','SubStage'],values = 'Value')

但这会产生:

KeyError: 'Level Stage not found'

是什么赋予了？

我正在创建一个 Fiverr.com 的克隆作为项目。

我的 base.html 中有一个带有类别标题的标题，如果我点击它，应该只过滤掉相关类别中的演出以进行显示。

发生的情况是，无论如何它总是重定向到主页。我做了一些测试，相信应该是因为KeyError，并且没有链接正确传递给函数。

代码如下：

视图.py

def category(request, link):

  categories = {
      "Graphics & Design": "GD",
      "Digital & Marketing": "DM",
      "Video & Animation": "VA",
      "Music & Audio": "MA",
      "Programming & Tech": "PT"
  }
  try:
      gigs = Gig.objects.filter(category=categories[link])
      return render(request, 'home.html', {"gigs": gigs})
  except KeyError:
      return redirect('home')

模型.py

class Gig(models.Model):
  CATEGORY_CHOICES = (
      ("GD", "Graphics & Design"),
      ("DM", "Digital & Marketing"),
      ("VA", "Video & Animation"),
      ("MA", "Music & Audio"),
      ("PT", "Programming & Tech")
  )
  title = …

我一直很难处理烧瓶中的会话。自从我在本地环境中管理应用程序以来，一切都运行良好，包括烧瓶会话。但是当我已经在渲染中托管它时，我总是在每条路线中收到此错误。

[55] [ERROR] Error handling request /valle-de-guadalupe
Traceback (most recent call last):
    File "/opt/render/project/src/.venv/lib/python3.7/site-packages/flask/app.py", line 2525, in wsgi_app
      response = self.full_dispatch_request()
    File "/opt/render/project/src/.venv/lib/python3.7/site-packages/flask/app.py", line 1822, in full_dispatch_request
      rv = self.handle_user_exception(e)
    File "/opt/render/project/src/.venv/lib/python3.7/site-packages/flask/app.py", line 1820, in full_dispatch_request
      rv = self.dispatch_request()
    File "/opt/render/project/src/.venv/lib/python3.7/site-packages/flask/app.py", line 1796, in dispatch_request
      return self.ensure_sync(self.view_functions[rule.endpoint])(**view_args)
    File "/opt/render/project/src/app_folder/routes/public.py", line 35, in valle_de_guadalupe
     return render_template("public/cities/valle_guadalupe.html")
    File "/opt/render/project/src/.venv/lib/python3.7/site-packages/flask/templating.py", line 147, in render_template
      return _render(app, template, context)
    File "/opt/render/project/src/.venv/lib/python3.7/site-packages/flask/templating.py", line 128, in _render
      app.update_template_context(context)
    File "/opt/render/project/src/.venv/lib/python3.7/site-packages/flask/app.py", line 994, in update_template_context
      context.update(func())
    File …