Suppose I have a dataframe like:
ticker MS AAPL
field price volume price volume
0 -0.861210 -0.319607 -0.855145 0.635594
1 -1.986693 -0.526885 -1.765813 1.696533
2 -0.154544 -1.152361 -1.391477 -2.016119
3 0.621641 -0.109499 0.143788 -0.050672
generated from following codes, please ignore the numbers which are just as examples
columns = pd.MultiIndex.from_tuples([('MS', 'price'), ('MS', 'volume'), ('AAPL', 'price'), ('AAPL', 'volume')], names=['ticker', 'field'])
data = np.random.randn(4, 4)
df = pd.DataFrame(data, columns=columns)
Now, I would like to calculate pct_change() or any function user definded on each price column, and add a new column on 'field' level to store the result.
I know how to do it elegantly if the data is a Panel, which is deprecated since ver 0.20. Suppose panel's 3 axis are date, ticker and field:
p[:,:, 'ret'] = p[:,:,'price'].pct_change()
That's all. But I have not found a similar elegant way to do it with multiple index dataframe.
You can using IndexSlice
df.loc[:,pd.IndexSlice[:,'price']].apply(pd.Series.pct_change).rename(columns={'price':'ret'})
Out[1181]:
ticker MS AAPL
field ret ret
0 NaN NaN
1 -1.420166 -0.279805
2 3.011155 0.062529
3 -1.609004 0.759954
def cstm(s):
return s.pct_change()
new = pd.concat(
[df.xs('price', 1, 1).apply(cstm)],
axis=1, keys=['new']
).swaplevel(0, 1, 1)
df.join(new).sort_index(1)
ticker AAPL MS
field new price volume new price volume
0 NaN -0.855145 0.635594 NaN -0.861210 -0.319607
1 1.064928 -1.765813 1.696533 1.306863 -1.986693 -0.526885
2 -0.211991 -1.391477 -2.016119 -0.922211 -0.154544 -1.152361
3 -1.103335 0.143788 -0.050672 -5.022430 0.621641 -0.109499
Or
def cstm(s):
return s.pct_change()
df.stack(0).assign(
new=lambda d: d.groupby('ticker').price.apply(cstm)
).unstack().swaplevel(0, 1, 1).sort_index(1)
Related
I have two dataframes. One is the basevales (df) and the other is an offset (df2).
How do I create a third dataframe that is the first dataframe offset by matching values (the ID) in the second dataframe?
This post doesn't seem to do the offset... Update only some values in a dataframe using another dataframe
import pandas as pd
# initialize list of lists
data = [['1092', 10.02], ['18723754', 15.76], ['28635', 147.87]]
df = pd.DataFrame(data, columns = ['ID', 'Price'])
offsets = [['1092', 100.00], ['28635', 1000.00], ['88273', 10.]]
df2 = pd.DataFrame(offsets, columns = ['ID', 'Offset'])
print (df)
print (df2)
>>> print (df)
ID Price
0 1092 10.02
1 18723754 15.76 # no offset to affect it
2 28635 147.87
>>> print (df2)
ID Offset
0 1092 100.00
1 28635 1000.00
2 88273 10.00 # < no match
This is want I want to produce: The price has been offset by matching
ID Price
0 1092 110.02
1 18723754 15.76
2 28635 1147.87
I've also looked at Pandas Merging 101
I don't want to add columns to the dataframe, and I don;t want to just replace column values with values from another dataframe.
What I want is to add (sum) column values from the other dataframe to this dataframe, where the IDs match.
The closest I come is df_add=df.reindex_like(df2) + df2 but the problem is that it sums all columns - even the ID column.
Try this :
df['Price'] = pd.merge(df, df2, on=["ID"], how="left")[['Price','Offset']].sum(axis=1)
What I need to do is to compare 2 fields in a row in a csv-file:
Data looks like this:
store;ean;price;retail_price;quantity
001;0888721396226;200;200;2
001;0888721396233;200;159;2
001;2194384654084;299;259;7
001;2194384654091;199.95;199.95;8
in case that "price" is equal to "retail_price" the field retail_price must be reduced by a given percent-value, e.g. -10%
so in the example data, the first and last line should be changed to 180 and 179,955
I´m completely new to pandas and after reading the "getting started" part I did not find anything that I could set upon ...
so any help or hint (just point me in the direction, I will fiddle it out myself then) is appreciated,
Kind regards!
Use Series.eq for compare both values and if same multiple retail_price by 0.9 else not in numpy.where:
mask = df['price'].eq(df['retail_price'])
df['retail_price'] = np.where(mask, df['retail_price'].mul(0.9), df['retail_price'])
print (df)
store ean price retail_price quantity
0 1 888721396226 200.00 180.000 2
1 1 888721396233 200.00 159.000 2
2 1 2194384654084 299.00 259.000 7
3 1 2194384654091 199.95 179.955 8
Or you can use DataFrame.loc for multiple only matched rows by 0.9:
mask = df['price'].eq(df['retail_price'])
df.loc[mask, 'retail_price'] *= 0.9
#working like
df.loc[mask, 'retail_price'] = df.loc[mask, 'retail_price'] * 0.9
EDIT: for filter rows not matched mask (with Falses in mask) use:
df2 = df[~mask].copy()
print (df2)
store ean price retail_price quantity
1 1 888721396233 200.0 159.0 2
2 1 2194384654084 299.0 259.0 7
print (mask)
0 True
1 False
2 False
3 True
dtype: bool
This ist my code:
import pandas as pd
import numpy as np
import sys
with open('prozente.txt', 'r') as f: #create multiplicator from static value in File "prozente.txt"
prozente = int(f.readline())
mulvalue = 1-(prozente/100)
df = pd.read_csv('1.csv', sep=';', header=1, names=['store','ean','price','retail_price','quantity'])
mask = df['price'].eq(df['retail_price'])
df['retail_price'] = np.where(mask, df['retail_price'].mul(mulvalue).round(2), df['retail_price'])
df2 = df[~mask].copy()
df.to_csv('output.csv', columns=['store','ean','price','retail_price','quantity'],sep=';', index=False)
print(df)
print(df2)
using this as 1.csv:
store;ean;price;retail_price;quantity
001;0888721396226;200;200;2
001;0888721396233;200;159;2
001;2194384654084;299;259;7
001;2194384654091;199.95;199.95;8
The content of file "prozente.txt" is
25
I have the following dataframe (time-series of returns truncated for succinctness):
import pandas as pd
import numpy as np
df = pd.DataFrame({'return':np.array([np.nan, np.nan, np.nan, 0.015, -0.024, 0.033, 0.021, 0.014, -0.092])})
I'm trying to start the index (i.e., "base-100") at the last NaN before the first return - while at the same time keep the NaNs preceding the 100 value in place - (thinking in terms of appending to existing dataframe and for graphing purposes).
I only have found a way to create said index when there are no NaNs in the return vector:
df['index'] = 100*np.exp(np.nan_to_num(df['return'].cumsum()))
Any ideas - thx in advance!
If your initial array is
zz = np.array([np.nan, np.nan, np.nan, 0.015, -0.024, 0.033, 0.021, 0.014, -0.092])
Then you can obtain your desired output like this (although there's probably a more optimized way to do it):
np.concatenate((zz[:np.argmax(np.isfinite(zz))],
100*np.exp(np.cumsum(zz[np.isfinite(zz)]))))
Use Series.isna, change order by indexing and get index of last NaN by Series.idxmax:
idx = df['return'].isna().iloc[::-1].idxmax()
Pass to DataFrame.loc, repalce missing value and use cumulative sum:
df['return'] = df.loc[idx:, 'return'].fillna(100).cumsum()
print (df)
return
0 NaN
1 NaN
2 100.000
3 100.015
4 99.991
5 100.024
6 100.045
7 100.059
8 99.967
You can use Series.isna with Series.cumsum and compare by max, then replace last NaN by Series.fillna and last use cumulative sum:
s = df['return'].isna().cumsum()
df['return'] = df['return'].mask(s.eq(s.max()), df['return'].fillna(100)).cumsum()
print (df)
return
0 NaN
1 NaN
2 100.000
3 100.015
4 99.991
5 100.024
6 100.045
7 100.059
8 99.967
I've got a pandas DataFrame filled mostly with real numbers, but there is a few nan values in it as well.
How can I replace the nans with averages of columns where they are?
This question is very similar to this one: numpy array: replace nan values with average of columns but, unfortunately, the solution given there doesn't work for a pandas DataFrame.
You can simply use DataFrame.fillna to fill the nan's directly:
In [27]: df
Out[27]:
A B C
0 -0.166919 0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3 NaN -2.027325 1.533582
4 NaN NaN 0.461821
5 -0.788073 NaN NaN
6 -0.916080 -0.612343 NaN
7 -0.887858 1.033826 NaN
8 1.948430 1.025011 -2.982224
9 0.019698 -0.795876 -0.046431
In [28]: df.mean()
Out[28]:
A -0.151121
B -0.231291
C -0.530307
dtype: float64
In [29]: df.fillna(df.mean())
Out[29]:
A B C
0 -0.166919 0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3 -0.151121 -2.027325 1.533582
4 -0.151121 -0.231291 0.461821
5 -0.788073 -0.231291 -0.530307
6 -0.916080 -0.612343 -0.530307
7 -0.887858 1.033826 -0.530307
8 1.948430 1.025011 -2.982224
9 0.019698 -0.795876 -0.046431
The docstring of fillna says that value should be a scalar or a dict, however, it seems to work with a Series as well. If you want to pass a dict, you could use df.mean().to_dict().
Try:
sub2['income'].fillna((sub2['income'].mean()), inplace=True)
In [16]: df = DataFrame(np.random.randn(10,3))
In [17]: df.iloc[3:5,0] = np.nan
In [18]: df.iloc[4:6,1] = np.nan
In [19]: df.iloc[5:8,2] = np.nan
In [20]: df
Out[20]:
0 1 2
0 1.148272 0.227366 -2.368136
1 -0.820823 1.071471 -0.784713
2 0.157913 0.602857 0.665034
3 NaN -0.985188 -0.324136
4 NaN NaN 0.238512
5 0.769657 NaN NaN
6 0.141951 0.326064 NaN
7 -1.694475 -0.523440 NaN
8 0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794
In [22]: df.mean()
Out[22]:
0 -0.251534
1 -0.040622
2 -0.841219
dtype: float64
Apply per-column the mean of that columns and fill
In [23]: df.apply(lambda x: x.fillna(x.mean()),axis=0)
Out[23]:
0 1 2
0 1.148272 0.227366 -2.368136
1 -0.820823 1.071471 -0.784713
2 0.157913 0.602857 0.665034
3 -0.251534 -0.985188 -0.324136
4 -0.251534 -0.040622 0.238512
5 0.769657 -0.040622 -0.841219
6 0.141951 0.326064 -0.841219
7 -1.694475 -0.523440 -0.841219
8 0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794
Although, the below code does the job, BUT its performance takes a big hit, as you deal with a DataFrame with # records 100k or more:
df.fillna(df.mean())
In my experience, one should replace NaN values (be it with Mean or Median), only where it is required, rather than applying fillna() all over the DataFrame.
I had a DataFrame with 20 variables, and only 4 of them required NaN values treatment (replacement). I tried the above code (Code 1), along with a slightly modified version of it (code 2), where i ran it selectively .i.e. only on variables which had a NaN value
#------------------------------------------------
#----(Code 1) Treatment on overall DataFrame-----
df.fillna(df.mean())
#------------------------------------------------
#----(Code 2) Selective Treatment----------------
for i in df.columns[df.isnull().any(axis=0)]: #---Applying Only on variables with NaN values
df[i].fillna(df[i].mean(),inplace=True)
#---df.isnull().any(axis=0) gives True/False flag (Boolean value series),
#---which when applied on df.columns[], helps identify variables with NaN values
Below is the performance i observed, as i kept on increasing the # records in DataFrame
DataFrame with ~100k records
Code 1: 22.06 Seconds
Code 2: 0.03 Seconds
DataFrame with ~200k records
Code 1: 180.06 Seconds
Code 2: 0.06 Seconds
DataFrame with ~1.6 Million records
Code 1: code kept running endlessly
Code 2: 0.40 Seconds
DataFrame with ~13 Million records
Code 1: --did not even try, after seeing performance on 1.6 Mn records--
Code 2: 3.20 Seconds
Apologies for a long answer ! Hope this helps !
If you want to impute missing values with mean and you want to go column by column, then this will only impute with the mean of that column. This might be a little more readable.
sub2['income'] = sub2['income'].fillna((sub2['income'].mean()))
# To read data from csv file
Dataset = pd.read_csv('Data.csv')
X = Dataset.iloc[:, :-1].values
# To calculate mean use imputer class
from sklearn.impute import SimpleImputer
imputer = SimpleImputer(missing_values=np.nan, strategy='mean')
imputer = imputer.fit(X[:, 1:3])
X[:, 1:3] = imputer.transform(X[:, 1:3])
Directly use df.fillna(df.mean()) to fill all the null value with mean
If you want to fill null value with mean of that column then you can use this
suppose x=df['Item_Weight'] here Item_Weight is column name
here we are assigning (fill null values of x with mean of x into x)
df['Item_Weight'] = df['Item_Weight'].fillna((df['Item_Weight'].mean()))
If you want to fill null value with some string then use
here Outlet_size is column name
df.Outlet_Size = df.Outlet_Size.fillna('Missing')
Pandas: How to replace NaN (nan) values with the average (mean), median or other statistics of one column
Say your DataFrame is df and you have one column called nr_items. This is: df['nr_items']
If you want to replace the NaN values of your column df['nr_items'] with the mean of the column:
Use method .fillna():
mean_value=df['nr_items'].mean()
df['nr_item_ave']=df['nr_items'].fillna(mean_value)
I have created a new df column called nr_item_ave to store the new column with the NaN values replaced by the mean value of the column.
You should be careful when using the mean. If you have outliers is more recommendable to use the median
Another option besides those above is:
df = df.groupby(df.columns, axis = 1).transform(lambda x: x.fillna(x.mean()))
It's less elegant than previous responses for mean, but it could be shorter if you desire to replace nulls by some other column function.
using sklearn library preprocessing class
from sklearn.impute import SimpleImputer
missingvalues = SimpleImputer(missing_values = np.nan, strategy = 'mean', axis = 0)
missingvalues = missingvalues.fit(x[:,1:3])
x[:,1:3] = missingvalues.transform(x[:,1:3])
Note: In the recent version parameter missing_values value change to np.nan from NaN
I use this method to fill missing values by average of a column.
fill_mean = lambda col : col.fillna(col.mean())
df = df.apply(fill_mean, axis = 0)
You can also use value_counts to get the most frequent values. This would work on different datatypes.
df = df.apply(lambda x:x.fillna(x.value_counts().index[0]))
Here is the value_counts api reference.
I am trying to create a row that has columns from t0 to t(n).
I have a complete data frame (df) that stores the full set of data, and a data series (df_t) specific time markers I am interested in.
What I want is to create a row that has the time marker as t0 then the previous [sequence_length] rows from the complete data frame.
def t_data(df, df_t, col_names, sequence_length):
df_ret = pd.DataFrame()
for i in range(sequence_length):
col_names_seq = [col_name + "_" + str(i) for col_name in col_names]
df_ret[col_names_seq] = df[df.shift(i)["time"].isin(df_t)][col_names]
return df_ret
Running:
t_data(df, df_t, ["close"], 3)
I get:
close_0 close_1 close_2
1110 1.32080 NaN NaN
2316 1.30490 NaN NaN
2549 1.30290 NaN NaN
The obvious line in issue is:
df[df.shift(i)["time"].isin(df_t)][col_names]
I have tried several ways but cant seem to select data surrounding a subset.
Sample (df):
time open close high low volume EMA21 EMA13 EMA9
20 2005-01-10 04:10:00 1.3071 1.3074 1.3075 1.3070 32.0 1.306624 1.306790 1.306887
21 2005-01-10 04:15:00 1.3074 1.3073 1.3075 1.3073 16.0 1.306685 1.306863 1.306969
22 2005-01-10 04:20:00 1.3073 1.3072 1.3074 1.3072 35.0 1.306732 1.306911 1.307015
Sample (df_t):
1110 2005-01-13 23:00:00
2316 2005-01-18 03:30:00
2549 2005-01-18 22:55:00
Name: time, dtype: datetime64[ns]
I don’t have data but hope this drawing helps:
def t_data(df, df_T, n):
# Get the indexs of the original df that matches with the values of df_T
indexs = df.reset_index().merge(df_T, how="inner")['index'].tolist()
#create new index list where we will store the index-n vales
newIndex = []
#create list of values to subtract from index
toSub = np.arange(n)
#loop over index values and subtract the values, and append in newIndex
for i in indexs:
for sub in toSub:
v = i - sub
newIndex.append(v)
#Use iloc to get all the rows in the original df with the newIndex values that we want
closedCosts = df.iloc[newIndex].reset_index(drop = True)["close"].values
#concat our data back to df_T, and reshape closedCosts by n columns
df_final = pd.concat([df_T, pd.DataFrame(closedCosts.reshape(-1, n))], axis= 1)
#return final df
return df_final
This should do what you're asking for. The easiest way to do this is to figure out all the indexs that you would want from the original df with its corresponding closing value. Note: you will have to rename the columns after this, but all the values are there.