Pandas loc is returning series not df - python

The following code returns a series for y when I want a df. Ultimately I am pulling rows out of a larger raw df (df) to create a smaller df (Cand) of results. I have created Cand as the new empty df to be populated.
Cand = pd.DataFrame(columns=['SR','Hits','Breaks'])
x = df.loc[df['Breaks'] == 0]
y = x.loc[x['Hits'].idxmax()]
Cand.append(y)
x is correctly reflected as a df, but y becomes a series and so does not populate Cand.
I have looked around but cannot find a similar problem. Thanks in advance.

Your issue would not be that you aren't passing a DataFrame to append(), but that .append() here is not in-place; try reassigning the return of append() to Cand as Cand = Cand.append(y), given that append returns your initial DataFrame + other (Cand + y, in this case).
Side Note:
You can return a DataFrame from .loc by using double square brackets.
Example: y = x.loc[[x['Hits'].idxmax()]]

Related

Create new column with reverse order using pandas dataframe

I need to create a new column X containing the reverse order value of x shown below.
x X
aa01 01aa
bb02 02bb
cc03 03cc
I did slice and concatenate them manually and it worked anyway, but I am looking for a "smarter" way doing this.
df["X1"] = df["x"].str.slice(0,2)
df["X2"] = df["x"].str.slice(2,4)
df['X'] = df["X2"]+ df["X1"].map(str)
Faster way would be using list comprehension instead of pandas str functions:
df['X'] = [s[-2:]+s[:2] for s in df.x]

Get all previous values for every row

I'm about to write a backtesting tool and so for every row I'd like to have access to all the dataframe till the given row. In the following example I'm doing it from a fixed index using a loop. I'm wondering if there is any better solution.
import numpy as np
import pandas as pd
N
df = pd.DataFrame({"a":np.arange(N)})
for i in range(3,N):
print(df["a"][:i].values)
UPDATE (toy example)
I need to apply a custom function to all the previous values. Here as a toy example I will use the sum of the square of all previous values.
def toyFun(v):
return np.sum(v**2)
res = np.empty(N)
res[:] = np.nan
for i in range(3, N):
res[i] = toyFun(df["a"][:i].values)
df["res"] = res
If you are indexing rows for a particular column say 'a', you can use .iloc indexer (i stands for index, loc means location) to index on the columns.
df = pd.DataFrame({'a': [1,2,3,4]})
print(df.a.iloc[:2]) # get first two values
So, you can do:
for i in range(3, 10):
print(df.a.iloc[:i])
The best way is to use a temporary column with the direct results, that way you are not re-calculating everything.
df["a"].apply(lambda x: x**2).cumsum()
Then re-index as you which:
res[3:] = df["a"].apply(lambda x: x**2).cumsum()[2:N-1].values
or directly to the dataframe.

Apply a custom function on columns in a pandas dataframe

I want to do something equivalent to
Select x,y,z from data where f(x, Y);
And f is my customized function that looks into the values of specific columns in a row and returns True or False. I tried the following:
df = df.ix[_is_detection_in_window(df['Product'], df['CreatedDate'])== True]
But I get
TypeError: 'Series' objects are mutable, thus they cannot be hashed
I think it does not iterate over the rows.
I also tried:
i = 0
for index, row in df.iterrows():
if _is_detection_in_window(row['Product'], row['CreatedDate']):
print 'in range '
new_df.iloc[i] = row
i+= 1
df = new_df
but I get :
IndexError: single positional indexer is out-of-bounds
It seems like your function doesn't accept Series, but that can be changed using np.vectorize:
v = np.vectorize(_is_detection_in_window)
df = df.loc[v(df['Product'], df['CreatedDate'])]
Furthermore, you should refrain from using .ix which is now deprecated as of v20.
Not sure how your function looks, but I assume it returns a list of bools equal to the number of rows in your df:
df = df.iloc[_is_detection_in_window(df['Product'], df['CreatedDate']), :]

Reading values from Pandas dataframe rows into equations and entering result back into dataframe

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')))

How to iterate over columns of pandas dataframe to run regression

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

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