我有一个名为“文件名”的 csv 文件，想要以 64float 的形式读取这些数据，但“小时”列除外。我使用 pd.read_csv - 函数和转换器来管理它。
df = pd.read_csv("../data/filename.csv",
delimiter = ';',
date_parser = ['hour'],
skiprows = 1,
converters={'column1': lambda x: float(x.replace ('.','').replace(',','.'))})
现在，我有两点:
第一:
分隔符与 ; 一起使用。 ，但是如果我在记事本中查看我的数据，有“，”，而不是“;”。但如果我采用 ',' 我得到: 'pandas.parser.CParserError: 标记数据时出错。 C 错误:第 13 行应有 7 个字段，但看到了 9'
第二:
如果我想对所有列使用转换器，我怎样才能得到这个？!什么是正确的术语？ 我尝试在读取函数中使用 dtype = float，但出现“AttributeError: 'NoneType' object has no attribute 'dtype'” 发生了什么？这就是为什么我想用转换器来管理它的原因。
数据:
,hour,PV,Wind onshore,Wind offshore,PV.1,Wind onshore.1,Wind offshore.1,PV.2,Wind onshore.2,Wind offshore.2 0,1,0.0,"12,985.0","9,614.0",0.0,"32,825.5","9,495.7",0.0,"13,110.3","10,855.5" 1,2,0.0,"12,908.9","9,290.8",0.0,"36,052.3","9,589.1",0.0,"13,670.2","10,828.6" 2,3,0.0,"12,740.9","8,886.9",0.0,"38,540.9","10,087.3",0.0,"14,610.8","10,828.6" 3,4,0.0,"12,485.3","8,644.5",0.0,"40,734.0","10,087.3",0.0,"15,638.3","10,343.7" 4,5,0.0,"11,188.5","8,079.0",0.0,"42,688.0","10,087.3",0.0,"16,809.4","10,343.7" 5,6,0.0,"11,219.0","7,594.2",0.0,"43,333.5","10,025.0",0.0,"18,266.9","10,343.7"
最佳答案
这应该有效:
In [40]:
# text data
temp=''',hour,PV,Wind onshore,Wind offshore,PV.1,Wind onshore.1,Wind offshore.1,PV.2,Wind onshore.2,Wind offshore.2
0,1,0.0,"12,985.0","9,614.0",0.0,"32,825.5","9,495.7",0.0,"13,110.3","10,855.5"
1,2,0.0,"12,908.9","9,290.8",0.0,"36,052.3","9,589.1",0.0,"13,670.2","10,828.6"
2,3,0.0,"12,740.9","8,886.9",0.0,"38,540.9","10,087.3",0.0,"14,610.8","10,828.6"
3,4,0.0,"12,485.3","8,644.5",0.0,"40,734.0","10,087.3",0.0,"15,638.3","10,343.7"
4,5,0.0,"11,188.5","8,079.0",0.0,"42,688.0","10,087.3",0.0,"16,809.4","10,343.7"
5,6,0.0,"11,219.0","7,594.2",0.0,"43,333.5","10,025.0",0.0,"18,266.9","10,343.7"'''
# so read the csv, pass params quotechar and the thousands character
df = pd.read_csv(io.StringIO(temp), quotechar='"', thousands=',')
df
Out[40]:
Unnamed: 0 hour PV Wind onshore Wind offshore PV.1 Wind onshore.1 \
0 0 1 0 12985.0 9614.0 0 32825.5
1 1 2 0 12908.9 9290.8 0 36052.3
2 2 3 0 12740.9 8886.9 0 38540.9
3 3 4 0 12485.3 8644.5 0 40734.0
4 4 5 0 11188.5 8079.0 0 42688.0
5 5 6 0 11219.0 7594.2 0 43333.5
Wind offshore.1 PV.2 Wind onshore.2 Wind offshore.2
0 9495.7 0 13110.3 10855.5
1 9589.1 0 13670.2 10828.6
2 10087.3 0 14610.8 10828.6
3 10087.3 0 15638.3 10343.7
4 10087.3 0 16809.4 10343.7
5 10025.0 0 18266.9 10343.7
In [41]:
# check the dtypes
df.info()
<class 'pandas.core.frame.DataFrame'>
Int64Index: 6 entries, 0 to 5
Data columns (total 11 columns):
Unnamed: 0 6 non-null int64
hour 6 non-null int64
PV 6 non-null float64
Wind onshore 6 non-null float64
Wind offshore 6 non-null float64
PV.1 6 non-null float64
Wind onshore.1 6 non-null float64
Wind offshore.1 6 non-null float64
PV.2 6 non-null float64
Wind onshore.2 6 non-null float64
Wind offshore.2 6 non-null float64
dtypes: float64(9), int64(2)
memory usage: 576.0 bytes
所以基本上你需要将 quotechar='"' 和 thousands=',' 参数传递给 read_csv 来实现你想要的，请参阅文档:http://pandas.pydata.org/pandas-docs/stable/generated/pandas.read_csv.html#pandas.read_csv
编辑
如果您想在导入后进行转换(当您可以预先完成时，这是一种浪费)，那么您可以对每个感兴趣的列执行此操作:
In [43]:
# replace the comma separator
df['Wind onshore'] = df['Wind onshore'].str.replace(',','')
# convert the type
df['Wind onshore'] = df['Wind onshore'].astype(np.float64)
df['Wind onshore'].dtype
Out[43]:
dtype('float64')
首先替换所有感兴趣列上的逗号分隔符，然后像这样调用 convert_objects 会更快:df.convert_objects(convert_numeric=True)
关于python - 每个转换器以浮点形式读取数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/27982526/