I'm using this code in order to groupby my data by year
df = pd.read_csv('../input/companies-info-wikipedia-2021/sparql_2021-11-03_22-25-45Z.csv')
df = pd.read_csv('../input/companies-info-wikipedia-2021/sparql_2021-11-03_22-25-45Z.csv')
df_duplicate_name = df[df.duplicated(['name'])]
df = df.drop_duplicates(subset='name').reset_index()
df = df.drop(['a','type','index'],axis=1).reset_index()
df = df[~df['foundation'].str.contains('[A-Za-z]', na=False)]
df = df.drop([140,214,220])
df['foundation'] = df['foundation'].fillna(0)
df['foundation'] = pd.to_datetime(df['foundation'])
df['foundation'] = df['foundation'].dt.year
df = df.groupby('foundation')
But as a result it does not group it by foundation values:
0 0 Deutsche EuroShop AG 1999 http://dbpedia.org/resource/Germany Investment in shopping centers http://dbpedia.org/resource/Real_property 4 2.964E9 1.25E9 2.241E8 8.04E7
1 1 Industry of Machinery and Tractors 1996 http://dbpedia.org/resource/Belgrade http://dbpedia.org/resource/Tractors http://dbpedia.org/resource/Agribusiness 4 4.648E7 0.0 30000.0 -€0.47 million
2 2 TelexFree Inc. 2012 http://dbpedia.org/resource/Massachusetts 99 http://dbpedia.org/resource/Multi-level_marketing 7 did not disclose did not disclose did not disclose did not disclose
3 3 (prev. Common Cents Communications Inc.) 2012 http://dbpedia.org/resource/United_States 99 http://dbpedia.org/resource/Multi-level_marketing 7 did not disclose did not disclose did not disclose did not disclose
4 4 Bionor Holding AS 1993 http://dbpedia.org/resource/Oslo http://dbpedia.org/resource/Health_care http://dbpedia.org/resource/Biotechnology 18 NOK 253 395 million NOK 203 320 million 1.09499E8 NOK 49 020 million
... ... ... ... ... ... ... ... ... ... ... ...
255 255 Ageas SA/NV 1990 http://dbpedia.org/resource/Belgium http://dbpedia.org/resource/Insurance http://dbpedia.org/resource/Financial_services 45000 1.0872E11 1.348E10 1.112E10 9.792E8
256 256 Sharp Corporation 1912 http://dbpedia.org/resource/Japan Televisions, audiovisual, home appliances, inf... http://dbpedia.org/resource/Consumer_electronics 52876 NaN NaN NaN NaN
257 257 Erste Group Bank AG 2008 Vienna, Austria Retail and commercial banking, investment and ... http://dbpedia.org/resource/Financial_services 47230 2.71983E11 1.96E10 6.772E9 1187000.0
258 258 Manulife Financial Corporation 1887 200 Asset management, Commercial banking, Commerci... http://dbpedia.org/resource/Financial_services 34000 750300000000 47200000000 39000000000 4800000000
259 259 BP plc 1909 London, England, UK http://dbpedia.org/resource/Natural_gas http://dbpedia.org/resource/Petroleum_industry
I also tried with making it again pd.to_datetime and sorting by dt.year - but still unsuccessful.
Column names:
Index(['index', 'name', 'foundation', 'location', 'products', 'sector',
'employee', 'assets', 'equity', 'revenue', 'profit'],
dtype='object')
#Ruslan you simply need to use a "sorting" command, not a "groupby" . You can achieve this generally in two ways:
myDF.sort_value(by='column_name' , ascending= 'true', inplace=true)
or, in case you need to set your column as index, you would need to do this:
myDF.index.name = 'column_name'
myDF.sort_index(ascending=True)
GroupBy is a totally different command, it is used to make actions after you group values by some criteria. Such as find sum, average , min, max of values, grouped-by some criteria.
pandas.DataFrame.sort_values
pandas.DataFrame.groupby
I think you're misunderstanding how groupby() works.
You can't do df = df.groupby('foundation'). groupby() does not return a new DataFrame. Instead, it returns a GroupBy, which is essentially just a mapping from value grouped-by to a dataframe containg the rows that all share that value for the specified column.
You can, for example, print how many rows are in each group with the following code:
groups = df.groupby('foundation')
for val, sub_df in groups:
print(f'{val}: {sub_df.shape[0]} rows')
I have a df "data" as below
Name Quality city
Tom High A
nick Medium B
krish Low A
Jack High A
Kevin High B
Phil Medium B
I want group it by city and a create a new columns based on the column "quality" and calculate avegare as below
city High Medium Low High_Avg Medium_AVG Low_avg
A 2 0 1 66.66 0 33.33
B 1 1 0 50 50 0
I tried with the below script and I know it is completely wrong.
data_average = data_df.groupby(['city'], as_index = False).count()
Get a count of the frequencies, divide the outcome by the sum across columns, and finally concatenate the datframes into one :
result = pd.crosstab(df.city, df.Quality)
averages = result.div(result.sum(1).array, axis=0).mul(100).round(2).add_suffix("_Avg")
#combine the dataframes
pd.concat((result, averages), axis=1)
Quality High Low Medium High_Avg Low_Avg Medium_Avg
city
A 2 1 0 66.67 33.33 0.00
B 1 0 2 33.33 0.00 66.67
I have dataframe with two rows as header(name and unit). is there a way to groupby dataframe with unit. what Iam trying to acheive is group by similar units and run analysis on them.
df = pd.read_csv('filename',header=[0.1])
Customer length height adress city
Name meter meter bldg name
A 10 20 1 Delhi
C 30 20 10 Delhi
B 20 40 19 Delhi
D 40 50 10 Delhi
i am trying to isolate the dataframe with units(second header)
for example:
length height
10 20
30 20
20 40
40 50
New to Python and looking for some help.
I would like to divide values in two different rows (part of the same column) and then insert a new column with the calculated value
City 2017-18 Item
0 Boston 100 Primary
1 Boston 200 Secondary
2 Boston 300 Tertiary
3 Boston 400 Nat'l average
4 Chicago 500 Primary
5 Chicago 600 Secondary
6 Chicago 700 Tertiary
7 Chicago 800 Nat'l average
On the above Dataframe, I am trying to divide a City's Primary, Secondary and Tertiary values respectively by the Nat'l average for that City. The resultant answer to be populated in a new column part of the same Dataframe. After calculation, the row with the label 'Nat'l average' need to be deleted.
Appreciate your help...
City 2014-15 Item New_column
0 Boston 100 Primary 100/400
1 Boston 200 Secondary 200/400
2 Boston 300 Tertiary 300/400
3 Chicago 500 Primary 500/800
4 Chicago 600 Secondary 600/800
5 Chicago 700 Tertiary 700/800
If mean value is always last per groups divide column by Series created by GroupBy.transform and GroupBy.last:
df['new'] = df['2017-18'].div(df.groupby('City')['2017-18'].transform('last'))
If not first filter values with averages and divide by Series.maping Series:
s = df[df['Item'] == "Nat'l average"].set_index('City')['2017-18']
df['new'] = df['2017-18'].div(df['City'].map(s))
And last filter out rows by boolean indexing:
df = df[df['Item'] != "Nat'l average"]
print (df)
City 2017-18 Item new
0 Boston 100 Primary 0.250
1 Boston 200 Secondary 0.500
2 Boston 300 Tertiary 0.750
4 Chicago 500 Primary 0.625
5 Chicago 600 Secondary 0.750
6 Chicago 700 Tertiary 0.875
Detail:
print (df['City'].map(s))
0 400
1 400
2 400
3 400
4 800
5 800
6 800
7 800
Name: City, dtype: int64
I would like to convert the following dataframe into a json .
df:
A sector B sector C sector
TTM Ratio -- 35.99 12.70 20.63 14.75 23.06
RRM Sales -- 114.57 1.51 5.02 1.00 4594.13
MQR book 1.48 2.64 1.02 2.46 2.73 2.74
TTR cash -- 14.33 7.41 15.35 8.59 513854.86
In order to do so by using the function df.to_json() I would need to have unique names in column and indices.
Therefore what I am looking for is to convert the column names into a row and have default column numbers . In short I would like the following output:
df:
0 1 2 3 4 5
A sector B sector C sector
TTM Ratio -- 35.99 12.70 20.63 14.75 23.06
RRM Sales -- 114.57 1.51 5.02 1.00 4594.13
MQR book 1.48 2.64 1.02 2.46 2.73 2.74
TTR cash -- 14.33 7.41 15.35 8.59 513854.86
Turning the column names into the first row so I can make the conversion correctly .
You could also use vstack in numpy:
>>> df
x y z
0 8 7 6
1 6 5 4
>>> pd.DataFrame(np.vstack([df.columns, df]))
0 1 2
0 x y z
1 8 7 6
2 6 5 4
The columns become the actual first row in this case.
Use assign by list of range and original column names:
print (range(len(df.columns)))
range(0, 6)
#for python2 list can be omit
df.columns = [list(range(len(df.columns))), df.columns]
Or MultiIndex.from_arrays:
df.columns = pd.MultiIndex.from_arrays([range(len(df.columns)), df.columns])
Also is possible use RangeIndex:
print (pd.RangeIndex(len(df.columns)))
RangeIndex(start=0, stop=6, step=1)
df.columns = pd.MultiIndex.from_arrays([pd.RangeIndex(len(df.columns)), df.columns])
print (df)
0 1 2 3 4 5
A sector B sector C sector
TTM Ratio -- 35.99 12.70 20.63 14.75 23.06
RRM Sales -- 114.57 1.51 5.02 1.00 4594.13
MQR book 1.48 2.64 1.02 2.46 2.73 2.74
TTR cash -- 14.33 7.41 15.35 8.59 513854.86