[英]Is there a way to create a dataframe from a dictionary with list of values?
我有一个字典myDict
,我想使用此myDict
创建一个数据框df
,如下所示:
myDict = {
1: [''],
2: ['07/19/2017', ' 10/18/2007', '12/20/2002','12/20/2002' ],
3: ['07/19/2017', ' 10/18/2007'],
4: ['12/13/1993'],
5: [''],
6: ['08/01/2007'],
7: ['04/23/2007'],
8: ['02/06/2007'],
9: ['02/06/2007'],
10: ['11/08/2001'],
11: [''],
12: [''],
13: ['12/20/2002']
}
df
ID Col1 Col2 Col3 Col4
1
2 07/19/2017 10/18/2007 12/20/2002 12/20/2002
3 07/19/2017 10/18/2007
4 12/13/1993
5
6 08/01/2007
7 04/23/2007
8 02/06/2007
9 02/06/2007
10 11/08/2001
11
12
13 12/20/2002
我如何做到这一点? 谢谢。
将所有内容放到函数中都无法正常工作...
def split_Date(df):
Dates1 = df.set_index('IDX')['Date'].to_dict()
dates = {}
for k, v in Dates1.items():
v = v.split(',')
dates[k] = [i for i in v]
dates = {k: sorted(v, key=lambda x: datetime.strptime(x.strip(), "%m/%d/%Y") if x != "" else x) for k, v in dates.items()}
df_dates = pd.DataFrame.from_dict(dates, orient="index").fillna('').rename_axis("IDX").rename(columns="Date{}".format).reset_index()
df = pd.merge(df, df_dates, on='IDX', how='inner', suffixes=('_chem', '_df'))
return df #Adding this doesn't make any difference
在函数外部运行此代码非常有效。 但是,这要求我每次有新的data
都要在所有行中更改myData
的值。 这不像具有功能那样高效
Dates1 = myData.set_index('IDX')['Date'].to_dict()
dates = {}
for k, v in Dates1.items():
v = v.split(',')
dates[k] = [i for i in v]
dates = {k: sorted(v, key=lambda x: datetime.strptime(x.strip(), "%m/%d/%Y")
if x != "" else x) for k, v in dates.items()}
df_dates = pd.DataFrame.from_dict(dates, orient="index").fillna('').rename_axis("IDX").rename(columns="Date{}".format).reset_index()
myData = pd.merge(myData, df_dates, on='IDX', how='inner', suffixes=('_chem', '_df'))
您可以使用pd.DataFrame.from_dict读取它,并通过orient参数将键设置为索引:
pd.DataFrame.from_dict(myDict, orient="index").fillna('')
# 0 1 2 3
#1
#2 07/19/2017 10/18/2007 12/20/2002 12/20/2002
#3 07/19/2017 10/18/2007
#4 12/13/1993
#5
#6 08/01/2007
# ...
要将键设置为单独的列,可以使用reset_index :
(pd.DataFrame.from_dict(myDict, orient="index")
.fillna('')
.rename_axis("ID")
.rename(columns="Col{}".format)
.reset_index())
# ID Col0 Col1 Col2 Col3
#0 1
#1 2 07/19/2017 10/18/2007 12/20/2002 12/20/2002
#2 3 07/19/2017 10/18/2007
#3 4 12/13/1993
#4 5
# ...
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