I have a pySpark dataframe, where I have null values that I want to replace - however the value to replace with is different for different groups.
My data looks like this (appologies, I dont have a way to past it as text):
For group A I want to replace the null values with -999;
while for group B, I want to replace the null value with 0.
Currently, I split the data into sections, then do a df = df.fillna(-999) .
Is there a more efficient way of doing it? in psudo-code I was thinking something along the line of df = df.where(col('group') == A).fillna(lit(-999)).where(col('group') == B).fillna(lit(0)) but ofcourse, this doesn't work.
You can use when:
from pyspark.sql import functions as F
# Loop over all the columns you want to fill
for col in ('Col1', 'Col2', 'Col3'):
# compute here conditions to fill using a value or another
fill_a = F.col(col).isNull() & (F.col('Group') == 'A')
fill_b = F.col(col).isNull() & (F.col('Group') == 'B')
# Fill the column based on the different conditions
# using nested `when` - `otherwise`.
#
# Do not forget to add the last `otherwise` with the original
# values if none of the previous conditions have been met
filled_col = (
F.when(fill_a, -999)
.otherwise(
F.when(fill_b, 0)
.otherwise(F.col(col))
)
)
# 'overwrite' the original column with the filled column
df = df.withColumn(col, filled_col)
Another possible option is to use coalesce for each column with a "filler" column holding the replacement values:
import pyspark.sql.functions as F
for c in ['Col1', 'Col2', 'Col3']:
df = df.withColumn(c, F.coalesce(c, F.when(F.col('group') == 'A', -999)
.when(F.col('group') == 'B', 0)))
I am removing a number of records in a pandas data frame which contains diverse combinations of NaN in the 4-columns frame. I have created a function called complete_cases to provide indexes of rows which met the following condition: all columns in the row are NaN.
I have tried this function below:
def complete_cases(dataframe):
indx = []
indx = [x for x in list(dataframe.index) \
if dataframe.loc[x, :].isna().sum() ==
len(dataframe.columns)]
return indx
I am wondering should this is optimal enough or there is a better way to do this.
Absolutely. All you need to do is
df.dropna(axis = 0, how = 'any', inplace = True)
This will remove all rows that have at least one missing value, and updates the data frame "inplace".
I'd recommend to use loc, isna, and any with 'columns' axis, like this:
df.loc[df.isna().any(axis='columns')]
This way you'll filter only the results like the complete.cases in R.
A possible solution:
Count the number of columns with "NA" creating a column to save it
Based on this new column, filter the rows of the data frame as you wish
Remove the (now) unnecessary column
It is possible to do it with a lambda function. For example, if you want to remove rows that have 10 "NA" values:
df['count'] = df.apply(lambda x: 0 if x.isna().sum() == 10 else 1, axis=1)
df = df[df.count != 0]
del df['count']
I'm initializing a DataFrame:
columns = ['Thing','Time']
df_new = pd.DataFrame(columns=columns)
and then writing values to it like this:
for t in df.Thing.unique():
df_temp = df[df['Thing'] == t] #filtering the df
df_new.loc[counter,'Thing'] = t #writing the filter value to df_new
df_new.loc[counter,'Time'] = dftemp['delta'].sum(axis=0) #summing and adding that value to the df_new
counter += 1 #increment the row index
Is there are better way to add new values to the dataframe each time without explicitly incrementing the row index with 'counter'?
If I'm interpreting this correctly, I think this can be done in one line:
newDf = df.groupby('Thing')['delta'].sum().reset_index()
By grouping by 'Thing', you have the various "t-filters" from your for-loop. We then apply a sum() to 'delta', but only within the various "t-filtered" groups. At this point, the dataframe has the various values of "t" as the indices, and the sums of the "t-filtered deltas" as a corresponding column. To get to your desired output, we then bump the "t's" into their own column via reset_index().
I have a dataframe. For each row of the dataframe: I need to read values from two column indexes, pass these values to a set of equations, enter the result of each equation into its own column index in the same row, go to the next row and repeat.
After reading the responses to similar questions I tried:
import pandas as pd
DF = pd.read_csv("...")
Equation_1 = f(x, y)
Equation_2 = g(x, y)
for index, row in DF.iterrows():
a = DF[m]
b = DF[n]
DF[p] = Equation_1(a, b)
DF[q] = Equation_2(a, b)
Rather than iterating over DF, reading and entering new values for each row, this codes iterates over DF and enters the same values for each row. I am not sure what I am doing wrong here.
Also, from what I have read it is actually faster to treat the DF as a NumPy array and perform the calculation over the entire array at once rather than iterating. Not sure how I would go about this.
Thanks.
Turns out that this is extremely easy. All that must be done is to define two variables and assign the desired columns to them. Then set "the row to be replaced" equivalent to the equation containing the variables.
Pandas already knows that it must apply the equation to every row and return each value to its proper index. I didn't realize it would be this easy and was looking for more explicit code.
e.g.,
import pandas as pd
df = pd.read_csv("...") # df is a large 2D array
A = df[0]
B = df[1]
f(A,B) = ....
df[3] = f(A,B)
# If your equations are simple enough, do operations column-wise in Pandas:
import pandas as pd
test = pd.DataFrame([[1,2],[3,4],[5,6]])
test # Default column names are 0, 1
test[0] # This is column 0
test.icol(0) # This is COLUMN 0-indexed, returned as a Series
test.columns=(['S','Q']) # Column names are easier to use
test #Column names! Use them column-wise:
test['result'] = test.S**2 + test.Q
test # results stored in DataFrame
# For more complicated stuff, try apply, as in Python pandas apply on more columns :
def toyfun(df):
return df[0]-df[1]**2
test['out2']=test[['S','Q']].apply(toyfun, axis=1)
# You can also define the column names when you generate the DataFrame:
test2 = pd.DataFrame([[1,2],[3,4],[5,6]],columns = (list('AB')))
I have this code using Pandas in Python:
all_data = {}
for ticker in ['FIUIX', 'FSAIX', 'FSAVX', 'FSTMX']:
all_data[ticker] = web.get_data_yahoo(ticker, '1/1/2010', '1/1/2015')
prices = DataFrame({tic: data['Adj Close'] for tic, data in all_data.iteritems()})
returns = prices.pct_change()
I know I can run a regression like this:
regs = sm.OLS(returns.FIUIX,returns.FSTMX).fit()
but how can I do this for each column in the dataframe? Specifically, how can I iterate over columns, in order to run the regression on each?
Specifically, I want to regress each other ticker symbol (FIUIX, FSAIX and FSAVX) on FSTMX, and store the residuals for each regression.
I've tried various versions of the following, but nothing I've tried gives the desired result:
resids = {}
for k in returns.keys():
reg = sm.OLS(returns[k],returns.FSTMX).fit()
resids[k] = reg.resid
Is there something wrong with the returns[k] part of the code? How can I use the k value to access a column? Or else is there a simpler approach?
for column in df:
print(df[column])
You can use iteritems():
for name, values in df.iteritems():
print('{name}: {value}'.format(name=name, value=values[0]))
This answer is to iterate over selected columns as well as all columns in a DF.
df.columns gives a list containing all the columns' names in the DF. Now that isn't very helpful if you want to iterate over all the columns. But it comes in handy when you want to iterate over columns of your choosing only.
We can use Python's list slicing easily to slice df.columns according to our needs. For eg, to iterate over all columns but the first one, we can do:
for column in df.columns[1:]:
print(df[column])
Similarly to iterate over all the columns in reversed order, we can do:
for column in df.columns[::-1]:
print(df[column])
We can iterate over all the columns in a lot of cool ways using this technique. Also remember that you can get the indices of all columns easily using:
for ind, column in enumerate(df.columns):
print(ind, column)
You can index dataframe columns by the position using ix.
df1.ix[:,1]
This returns the first column for example. (0 would be the index)
df1.ix[0,]
This returns the first row.
df1.ix[:,1]
This would be the value at the intersection of row 0 and column 1:
df1.ix[0,1]
and so on. So you can enumerate() returns.keys(): and use the number to index the dataframe.
A workaround is to transpose the DataFrame and iterate over the rows.
for column_name, column in df.transpose().iterrows():
print column_name
Using list comprehension, you can get all the columns names (header):
[column for column in df]
Based on the accepted answer, if an index corresponding to each column is also desired:
for i, column in enumerate(df):
print i, df[column]
The above df[column] type is Series, which can simply be converted into numpy ndarrays:
for i, column in enumerate(df):
print i, np.asarray(df[column])
I'm a bit late but here's how I did this. The steps:
Create a list of all columns
Use itertools to take x combinations
Append each result R squared value to a result dataframe along with excluded column list
Sort the result DF in descending order of R squared to see which is the best fit.
This is the code I used on DataFrame called aft_tmt. Feel free to extrapolate to your use case..
import pandas as pd
# setting options to print without truncating output
pd.set_option('display.max_columns', None)
pd.set_option('display.max_colwidth', None)
import statsmodels.formula.api as smf
import itertools
# This section gets the column names of the DF and removes some columns which I don't want to use as predictors.
itercols = aft_tmt.columns.tolist()
itercols.remove("sc97")
itercols.remove("sc")
itercols.remove("grc")
itercols.remove("grc97")
print itercols
len(itercols)
# results DF
regression_res = pd.DataFrame(columns = ["Rsq", "predictors", "excluded"])
# excluded cols
exc = []
# change 9 to the number of columns you want to combine from N columns.
#Possibly run an outer loop from 0 to N/2?
for x in itertools.combinations(itercols, 9):
lmstr = "+".join(x)
m = smf.ols(formula = "sc ~ " + lmstr, data = aft_tmt)
f = m.fit()
exc = [item for item in x if item not in itercols]
regression_res = regression_res.append(pd.DataFrame([[f.rsquared, lmstr, "+".join([y for y in itercols if y not in list(x)])]], columns = ["Rsq", "predictors", "excluded"]))
regression_res.sort_values(by="Rsq", ascending = False)
I landed on this question as I was looking for a clean iterator of columns only (Series, no names).
Unless I am mistaken, there is no such thing, which, if true, is a bit annoying. In particular, one would sometimes like to assign a few individual columns (Series) to variables, e.g.:
x, y = df[['x', 'y']] # does not work
There is df.items() that gets close, but it gives an iterator of tuples (column_name, column_series). Interestingly, there is a corresponding df.keys() which returns df.columns, i.e. the column names as an Index, so a, b = df[['x', 'y']].keys() assigns properly a='x' and b='y'. But there is no corresponding df.values(), and for good reason, as df.values is a property and returns the underlying numpy array.
One (inelegant) way is to do:
x, y = (v for _, v in df[['x', 'y']].items())
but it's less pythonic than I'd like.
Most of these answers are going via the column name, rather than iterating the columns directly. They will also have issues if there are multiple columns with the same name. If you want to iterate the columns, I'd suggest:
for series in (df.iloc[:,i] for i in range(df.shape[1])):
...
assuming X-factor, y-label (multicolumn):
columns = [c for c in _df.columns if c in ['col1', 'col2','col3']] #or '..c not in..'
_df.set_index(columns, inplace=True)
print( _df.index)
X, y = _df.iloc[:,:4].values, _df.index.values