My dataframe until now,
and I am trying to convert cols which is a list of all columns from 0 to 188 ( cols = list(hdata.columns[ range(0,188) ]) ) which are in this format yyyy-mm to datetimeIndex. There are other few columns as well which are 'string' Names and can't be converted to dateTime hence,so I tried doing this,
hdata[cols].columns = pd.to_datetime(hdata[cols].columns) #convert columns to **datetimeindex**
But this is not working.
Can you please figure out what is wrong here?
Edit:
A better way to work on this type of data is to use Split-Apply-Combine method.
Step 1: Split the data which you want to perform some specific operation.
nonReqdf = hdata.iloc[:,188:].sort_index()
reqdf= reqdf.drop(['CountyName','Metro','RegionID','SizeRank'],axis=1)
Step 2: do the operations. In my case it was converting the dataframe columns with year and months to datetimeIndex. And resample it quarterly.
reqdf.columns = pd.to_datetime(reqdf.columns)
reqdf = reqdf.resample('Q',axis=1).mean()
reqdf = reqdf.rename(columns=lambda x: str(x.to_period('Q')).lower()).sort_index() # renaming so that string is yyyy**q**<1/2/3/4> like 2012q1 or 2012q2 likewise
Step 3: Combine the two splitted dataframe.(merge can be used but may depend on what you want)
reqdf = pd.concat([reqdf,nonReqdf],axis=1)
In order to modify some of the labels from an Index (be it for rows or columns), you need to use df.rename as in
for i in range(188):
df.rename({df.columns[i]: pd.to_datetime(df.columns[i])},
axis=1, inplace=True)
Or you can avoid looping by building a full sized index to cover all the columns with
df.columns = (
pd.to_datetime(cols) # pass the list with strings to get a partial DatetimeIndex
.append(df.columns.difference(cols)) # complete the index with the rest of the columns
)
Related
The first df I have is one that has station codes and names, along with lat/long (not as relevant), like so:
code name latitude longitude
I have another df with start/end dates for travel times. This df has only the station code, not the station name, like so:
start_date start_station_code end_date end_station_code duration_sec
I am looking to add columns that have the name of the start/end stations to the second df by matching the first df "code" and second df "start_station_code" / "end_station_code".
I am relatively new to pandas, and was looking for a way to optimize doing this as my current method takes quite a while. I use the following code:
for j in range(0, len(df_stations)):
for i in range(0, len(df)):
if(df_stations['code'][j] == df['start_station_code'][i]):
df['start_station'][i] = df_stations['name'][j]
if(df_stations['code'][j] == df['end_station_code'][i]):
df['end_station'][i] = df_stations['name'][j]
I am looking for a faster method, any help is appreciated. Thank you in advance.
Use merge. If you are familiar with SQL, merge is equivalent to LEFT JOIN:
cols = ["code", "name"]
result = (
second_df
.merge(first_df[cols], left_on="start_station_code", right_on="code")
.merge(first_df[cols], left_on="end_station_code", right_on="code")
.rename(columns={"code_x": "start_station_code", "code_y": "end_station_code"})
)
The answer by #Code-Different is very nearly correct. However the columns to be renamed are the name columns not the code columns. For neatness you will likely want to drop the additional code columns that get created by the merges. Using your names for the dataframes df and df_station the code needed to produce df_required is:
cols = ["code", "name"]
required_df = (
df
.merge(df_stations[cols], left_on="start_station_code", right_on="code")
.merge(df_stations[cols], left_on="end_station_code", right_on="code")
.rename(columns={"name_x": "start_station", "name_y": "end_station"})
.drop(columns = ['code_x', 'code_y'])
)
As you may notice the merge means that the dataframe acquires duplicate 'code' columns which get suffixed automatically, this is a built in default of the merge command. See https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.merge.html for more detail.
my df columns names are dates in this format: dd-mm-yy. when I use sort_index(axis = 1) it sort by the first two digits (which specify the days) so it doesn't make sense chronologically. How can I sort it automatically by taking into account also the months?
my df headers:
submitted_at 06-05-18 13-05-18 29-04-18
I expected the output of:
submitted_at 29-04-18 06-05-18 13-05-18
Convert the columns to datetime and use argsort to find the correct ordering. This will put all non-dates to the left in the order they occur, followed by the sorted dates.
import pandas as pd
df = pd.DataFrame(columns=['submitted_at', '06-05-18', '13-05-18', '29-04-18'])
idx = pd.to_datetime(df.columns, errors='coerce', format='%d-%m-%y').argsort()
df.iloc[:, idx]
Empty DataFrame
Columns: [submitted_at, 29-04-18, 06-05-18, 13-05-18]
Converting strings to datetime then sorting them with something like this :
from datetime import datetime
cols_as_date = [datetime.strptime(x,'%d-%m-%Y') for x in df.columns]
df = df[sorted(cols_as_data)]
just convert to DateTime your column
df['newdate']=pd.to_datetime(df.date,format='%d-%m-%y')
and then sort it using sort_values
df.sort_values(by='newdate')
I have a dataframe which can be generated from the code as given below
df = pd.DataFrame({'person_id' :[1,2,3],'date1':
['12/31/2007','11/25/2009','10/06/2005'],'val1':
[2,4,6],'date2': ['12/31/2017','11/25/2019','10/06/2015'],'val2':[1,3,5],'date3':
['12/31/2027','11/25/2029','10/06/2025'],'val3':[7,9,11]})
I followed the below solution to convert it from wide to long
pd.wide_to_long(df, stubnames=['date', 'val'], i='person_id',
j='grp').sort_index(level=0)
Though this works with sample data as shown below, it doesn't work with my real data which has more than 200 columns. Instead of person_id, my real data has subject_ID which is values like DC0001,DC0002 etc. Does "I" always have to be numeric? Instead it adds the stub values as new columns in my dataset and has zero rows
This is how my real columns looks like
My real data might contains NA's as well. So do I have to fill them with default values for wide_to_long to work?
Can you please help as to what can be the issue? Or any other approach to achieve the same result is also helpful.
Try adding additional argument in the function which allows the strings suffix.
pd.long_to_wide(.......................,suffix='\w+')
The issue is with your column names, the numbers used to convert from wide to long need to be at the end of your column names or you need to specify a suffix to groupby. I think the easiest solution is to create a function that accepts regex and the dataframe.
import pandas as pd
import re
def change_names(df, regex):
# Select one of three column groups
old_cols = df.filter(regex = regex).columns
# Create list of new column names
new_cols = []
for col in old_cols:
# Get the stubname of the original column
stub = ''.join(re.split(r'\d', col))
# Get the time point
num = re.findall(r'\d+', col) # returns a list like ['1']
# Make new column name
new_col = stub + num[0]
new_cols.append(new_col)
# Create dictionary mapping old column names to new column names
dd = {oc: nc for oc, nc in zip(old_cols, new_cols)}
# Rename columns
df.rename(columns = dd, inplace = True)
return df
tdf = pd.DataFrame({'person_id' :[1,2,3],'h1date': ['12/31/2007','11/25/2009','10/06/2005'],'t1val': [2,4,6],'h2date': ['12/31/2017','11/25/2019','10/06/2015'],'t2val':[1,3,5],'h3date': ['12/31/2027','11/25/2029','10/06/2025'],'t3val':[7,9,11]})
# Change date columns
tdf = change_names(tdf, 'date$')
tdf = change_names(tdf, 'val$')
print(tdf)
person_id hdate1 tval1 hdate2 tval2 hdate3 tval3
0 1 12/31/2007 2 12/31/2017 1 12/31/2027 7
1 2 11/25/2009 4 11/25/2019 3 11/25/2029 9
2 3 10/06/2005 6 10/06/2015 5 10/06/2025 11
This is quite late to answer this question. But putting the solution here in case someone else find it useful
tdf = pd.DataFrame({'person_id' :[1,2,3],'h1date': ['12/31/2007','11/25/2009','10/06/2005'],'t1val': [2,4,6],'h2date': ['12/31/2017','11/25/2019','10/06/2015'],'t2val':[1,3,5],'h3date': ['12/31/2027','11/25/2029','10/06/2025'],'t3val':[7,9,11]})
## You can use m13op22 solution to rename your columns with numeric part at the
## end of the column name. This is important.
tdf = tdf.rename(columns={'h1date': 'hdate1', 't1val': 'tval1',
'h2date': 'hdate2', 't2val': 'tval2',
'h3date': 'hdate3', 't3val': 'tval3'})
## Then use the non-numeric portion, (in this example 'hdate', 'tval') as
## stubnames. The mistake you were doing was using ['date', 'val'] as stubnames.
df = pd.wide_to_long(tdf, stubnames=['hdate', 'tval'], i='person_id', j='grp').sort_index(level=0)
print(df)
I set up a pandas dataframes that besides my data stores the respective units with it using a MultiIndex like this:
Name Relative_Pressure Volume_STP
Unit - ccm/g
Description p/p0
0 0.042691 29.3601
1 0.078319 30.3071
2 0.129529 31.1643
3 0.183355 31.8513
4 0.233435 32.3972
5 0.280847 32.8724
Now I can for example extract only the Volume_STP data by
Unit ccm/g
Description
0 29.3601
1 30.3071
2 31.1643
3 31.8513
4 32.3972
5 32.8724
With .values I can obtain a numpy array of the data. However how can I get the stored unit? I can't figure out what I need to do to receive the stored ccm/g string.
EDIT: Added example how data frame is generated
Let's say I have a string that looks like this:
Relative Volume # STP
Pressure
cc/g
4.26910e-02 29.3601
7.83190e-02 30.3071
1.29529e-01 31.1643
1.83355e-01 31.8513
2.33435e-01 32.3972
2.80847e-01 32.8724
3.34769e-01 33.4049
3.79123e-01 33.8401
I then use this function:
def read_result(contents, columns, units, descr):
df = pd.read_csv(StringIO(contents), skiprows=4, delim_whitespace=True,index_col=False,header=None)
df.drop(df.index[-1], inplace=True)
index = pd.MultiIndex.from_arrays((columns, units, descr))
df.columns = index
df.columns.names = ['Name','Unit','Description']
df = df.apply(pd.to_numeric)
return df
like this
def isotherm(contents):
columns = ['Relative_Pressure','Volume_STP']
units = ['-','ccm/g']
descr = ['p/p0','']
df = read_result(contents, columns, units, descr)
return df
to generate the DataFrame at the beginning of my question.
As df has a MultiIndex as columns, df.Volume_STP is still a pandas DataFrame. So you can still access its columns attribute, and the relevant item will be at index 0 because the dataframe contains only 1 Series.
So, you can extract the names that way:
print(df.Volume_STP.columns[0])
which should give: ('ccm/g', '')
At the end you extract the unit with .colums[0][0] and the description with .columns[0][1]
You can do something like this:
df.xs('Volume_STP', axis=1).columns.remove_unused_levels().get_level_values(0).tolist()[0]
Output:
'ccm/g'
Slice the dataframe from the 'Volume_STP' using xs, then select the columns remove the unused parts of the column headers, then get the value for the top most level of that slice which is the Units. Convert to a list as select the first value.
A generic way of accessing values on multi-index/columns is by using the index.get_level_values or columns.get_level_values functions of a data frame.
In your example, try df.columns.get_level_values(1) to access the second level of the multi-level column "Unit". If you have already selected a column, say "Volume_STP", then you have removed the top level and in this case, your units would be in the 0th level.
I have two date columns namely date1 and date2.
I am trying to select rows which have date1 later than date2
I tried to
print df[df.loc[df['date1']>df['date2']]]
but I recieved an error
ValueError: Boolean array expected for the condition, not float64
In either case, the idea is to retrieve a boolean mask. This boolean mask will then be used to index into the dataframe and retrieve corresponding rows. First, generate a mask:
mask = df['date1'] > df['date2']
Now, use this mask to index df:
df = df.loc[mask]
This can be written in a single line.
df = df.loc[df['date1'] > df['date2']]
You do not need to perform another level of indexing after this, df now has your final result. I recommend loc if you are planning to perform operations and reassignment on this filtered dataframe, because loc always returns a copy, while plain indexing returns a view.
Below are some more methods of doing the same thing:
Option 1
df.query
df.query('date1 > date2')
Option 2
df.eval
df[df.eval('date1 > date2')]
If your columns are not dates, you might as well cast them now. Use pd.to_datetime:
df.date1 = pd.to_datetime(df.date1)
df.date2 = pd.to_datetime(df.date2)
Or, when loading your CSV, make sure to set the parse_dates switch on:
df = pd.read_csv(..., parse_dates=['date1, date2'])