In my data frame every entry is a string which consists of at least one number. Sometimes there are multiple and identical entries in one cell.
data = {'INTERVAL': ['0,60', '0,8 0,8', '0,5 0,5 0,5']}
df = pd.DataFrame(data)
print(df)
How can I extract the value as floating number and replace the original column with the new simplified representation? I've tried to use the extract
df['INTERVAL'].str.extract('((\d+))')
command, however I failed.
Thank you in advance
This seems to work for me -
floats = df['INTERVAL'].str.extract("(^[0-9]*,[0-9]*) ?.*")
df['INTERVAL'] = floats[0].str.replace(",",".").astype(float)
Related
I have two Dataframes, one containing my data read in from a CSV file and another that has the data grouped by all of the columns but the last and reindexed to contain a column for the count of the size of the groups.
df_k1 = pd.read_csv(filename, sep=';')
columns_for_groups = list(df_k1.columns)[:-1]
k1_grouped = df_k1.groupby(columns_for_groups).size().reset_index(name="Count")
I need to create a series such that every row(i) in the series corresponds to row(i) in my original Dataframe but the contents of the series need to be the size of the group that the row belongs to in the grouped Dataframe. I currently have this, and it works for my purposes, but I was wondering if anyone knew of a faster or more elegant solution.
size_by_row = []
for row in df_k1.itertuples():
for group in k1_grouped.itertuples():
if row[1:-1] == group[1:-1]:
size_by_row.append(group[-1])
break
group_size = pd.Series(size_by_row)
I have a dataframe of monthly returns for 1,000 stocks with ids as column names.
monthly returns
I need to select only the columns that match the values in another dataframe which includes the ids I want.
permno list
I'm sure this is really quite simple, but I have been struggling for 2 days and if someone has an easy solution it would be so very much appreciated. Thank you.
You could convert the single-column permno list dataframe (osr_curr_permnos) into a list, and then use that list to select certain columns from your main dataframe (all_rets).
To convert the osr_curr_permnos column "0" into a list, you can use .to_list()
Then, you can use that list to slice all_rets and .copy() to make a fresh copy of it into a new dataframe.
The python code might look something like:
keep = osr_curr_permnos['0'].to_list()
selected_rets = all_rets[keep].copy()
"keep" would be a list, and "selected_rets" would be your new dataframe.
If there's a chance that osr_curr_permnos would have duplicates, you'll want to filter those out:
keep = osr_curr_permnos['0'].drop_duplicates().to_list()
selected_rets = all_rets[keep].copy()
As I expected, the answer was more simple than I was making it. Basically, I needed to take the integer values in my permnos list and recast those as strings.
osr_curr_permnos['0'] = osr_curr_permnos['0'].apply(str)
keep = osr_curr_permnos['0'].values
Then I can use that to select columns from my returns dataframe which had string values as column headers.
all_rets[keep]
It was all just a mismatch of int vs. string.
I have read a fixed width file and created a dataframe.
I have a field called claim number which is of length 15.In the data frame I see this field appearing as "1.902431e+14" rather than full 15 length claim number.
how can I resolve this so that I can see entire 15 length of claim number in data frame ?
For example, use pandas float_format option as follows:
#data claim_number dictionary
dictionary = {'claim_number': [1.902431111111141]}
#specification of claim number format
pd.options.display.float_format = '{:,.15f}'.format
#Create dataframe
df = pd.DataFrame(data=dictionary)
If you want to apply your format specifically to one column only, you can use style.format instead of pd.options.float_format as follows:
#data claim_number dictionary
dictionary = {'claim_number_1': [1.902431111111141], 'claim_number_2': ['some_string'], 'claim_number_3': [0.2323]}
#Create dataframe
df = pd.DataFrame(data=dictionary)
#style format single column
df.style.format({'claim_number_1': "{:,.15f}"})
More options on how to use style.format can be found here https://pandas.pydata.org/pandas-docs/stable/user_guide/style.html
Say I have a DataFrame with a column of Float64s, I'd like to group the dataframe by binning that column. I hear the cut function might help, but it's not defined over dataframes. Some work has been done (https://gist.github.com/tautologico/3925372), but I'd rather use a library function rather than copy-pasting code from the Internet. Pointers?
EDIT Bonus karma for finding a way of doing this by month over UNIX timestamps :)
You could bin dataframes based on a column of Float64s like this. Here my bins are increments of 0.1 from 0.0 to 1.0, binning the dataframe based on a column of 100 random numbers between 0.0 and 1.0.
using DataFrames #load DataFrames
df = DataFrame(index = rand(Float64,100)) #Make a DataFrame with some random Float64 numbers
df_array = map(x->df[(df[:index] .>= x[1]) .& (df[:index] .<x[2]),:],zip(0.0:0.1:0.9,0.1:0.1:1.0)) #Map an anonymous function that gets every row between two numbers specified by a tuple called x, and map that anonymous function to an array of tuples generated using the zip function.
This will produce an array of 10 dataframes, each one with a different 0.1-sized bin.
As for the UNIX timestamp question, I'm not as familiar with that side of things, but after playing around a bit maybe something like this could work:
using Dates
df = DataFrame(unixtime = rand(1E9:1:1.1E9,100)) #Make a dataframe with floats containing pretend unix time stamps
df[:date] = Dates.unix2datetime.(df[:unixtime]) #convert those timestamps to DateTime types
df[:year_month] = map(date->string(Dates.Year.(date))*" "*string(Dates.Month.(date)),df[:date]) #Make a string for every month in your time range
df_array = map(ym->df[df[:year_month] .== ym,:],unique(df[:year_month])) #Bin based on each unique year_month string
I am trying to preset the dimensions of my data frame in pandas so that I can have 500 rows by 300 columns. I want to set it before I enter data into the dataframe.
I am working on a project where I need to take a column of data, copy it, shift it one to the right and shift it down by one row.
I am having trouble with the last row being cut off when I shift it down by one row (eg: I started with 23 rows and it remains at 23 rows despite the fact that I shifted down by one and should have 24 rows).
Here is what I have done so far:
bolusCI = pd.DataFrame()
##set index to very high number to accommodate shifting row down by 1
bolusCI = bolus_raw[["Activity (mCi)"]].copy()
activity_copy = bolusCI.shift(1)
activity_copy
pd.concat([bolusCI, activity_copy], axis =1)
Thanks!
There might be a more efficient way to achieve what you are looking to do, but to directly answer your question you could do something like this to init the DataFrame with certain dimensions
pd.DataFrame(columns=range(300),index=range(500))
You just need to define the index and columns in the constructor. The simplest way is to use pandas.RangeIndex. It mimics np.arange and range in syntax. You can also pass a name parameter to name it.
pd.DataFrame
pd.Index
df = pd.DataFrame(
index=pd.RangeIndex(500),
columns=pd.RangeIndex(300)
)
print(df.shape)
(500, 300)