I have a data frame with a column called "Date"
and want all the values from this column to have the same value (the year only). Example:
City Date
Paris 01/04/2004
Lisbon 01/09/2004
Madrid 2004
Pekin 31/2004
What I want is:
City Date
Paris 2004
Lisbon 2004
Madrid 2004
Pekin 2004
Here is my code:
fr61_70xls = pd.ExcelFile('AMADEUS FRANCE 1961-1970.xlsx')
#Here we import the individual sheets and clean the sheets
years=(['1961','1962','1963','1964','1965','1966','1967','1968','1969','1970'])
fr={}
header=(['City','Country','NACE','Cons','Last_year','Op_Rev_EUR_Last_avail_yr','BvD_Indep_Indic','GUO_Name','Legal_status','Date_of_incorporation','Legal_status_date'])
for year in years:
# save every sheet in variable fr['1961'], fr['1962'] and so on
fr[year]=fr61_70xls.parse(year,header=0,parse_cols=10)
fr[year].columns=header
# drop the entire Legal status date column
fr[year]=fr[year].drop(['Legal_status_date','Date_of_incorporation'],axis=1)
# drop every row where GUO Name is empty
fr[year]=fr[year].dropna(axis=0,how='all',subset=[['GUO_Name']])
fr[year]=fr[year].set_index(['GUO_Name','Date_of_incorporation'])
It happens that in my DataFrames, called for example fr['1961']
the values of Date_of_incorporation
can be anything (strings, integer, and so on), so maybe it would be best to completely erase this column and then attach another column with only the year to the DataFrames?
replace() function is used to replace values in column (one value with another value on all columns). This method takes to_replace, value, inplace, limit, regex and method as parameters and returns a new DataFrame. When inplace=True is used, it replaces on existing DataFrame object and returns None value.
Suppose that you want to replace multiple values with multiple new values for an individual DataFrame column. In that case, you may use this template: df['column name'] = df['column name']. replace(['1st old value', '2nd old value', ...], ['1st new value', '2nd new value', ...])
To replace multiple values in a DataFrame, you can use DataFrame. replace() method with a dictionary of different replacements passed as argument.
As @DSM points out, you can do this more directly using the vectorised string methods:
df['Date'].str[-4:].astype(int)
Or using extract (assuming there is only one set of digits of length 4 somewhere in each string):
df['Date'].str.extract('(?P<year>\d{4})').astype(int)
An alternative slightly more flexible way, might be to use apply
(or equivalently map
) to do this:
df['Date'] = df['Date'].apply(lambda x: int(str(x)[-4:]))
# converts the last 4 characters of the string to an integer
The lambda function, is taking the input from the Date
and converting it to a year.
You could (and perhaps should) write this more verbosely as:
def convert_to_year(date_in_some_format):
date_as_string = str(date_in_some_format) # cast to string
year_as_string = date_in_some_format[-4:] # last four characters
return int(year_as_string)
df['Date'] = df['Date'].apply(convert_to_year)
Perhaps 'Year' is a better name for this column...
You can do a column transformation by using apply
Define a clean function to remove the dollar and commas and convert your data to float.
def clean(x):
x = x.replace("$", "").replace(",", "").replace(" ", "")
return float(x)
Next, call it on your column like this.
data['Revenue'] = data['Revenue'].apply(clean)
Or if one want to use lambda
function in the apply
function:
data['Revenue']=data['Revenue'].apply(lambda x:float(x.replace("$","").replace(",", "").replace(" ", "")))
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