The functionality of reindexing in python pandas can also be done python Series as below.
import pandas as pd
order = ['a','c','b']
series_data = pd.Series([1,2,3],index=order)
series_data
In that case why do we explicitly go for reindex?
Let's take an example using index available in Series
s = pd.Series([1,2,3], index=['k','f','t'])
s
# k 1
# f 2
# t 3
# dtype: int64
We can state that above series got assigned index with a datatype of int64.
Now let's proceed with reindex:
order = ['k','c','b']
s.reindex(order)
# k 1.0
# c NaN
# b NaN
# dtype: float64
As you can observe we passed two new index c and b which were not there in original series, so those values are assigned equal to NaN. Since NaN has dtype of float64, hence a final series results into only three indexes k, c and b with dtype as float64.
I hope this clears how index inside Series is different from reindex outside.
You can refer below link to understand about reindexing.
https://www.tutorialspoint.com/python_pandas/python_pandas_reindexing.htm
Related
I have the following data frame in IPython, where each row is a single stock:
In [261]: bdata
Out[261]:
<class 'pandas.core.frame.DataFrame'>
Int64Index: 21210 entries, 0 to 21209
Data columns:
BloombergTicker 21206 non-null values
Company 21210 non-null values
Country 21210 non-null values
MarketCap 21210 non-null values
PriceReturn 21210 non-null values
SEDOL 21210 non-null values
yearmonth 21210 non-null values
dtypes: float64(2), int64(1), object(4)
I want to apply a groupby operation that computes cap-weighted average return across everything, per each date in the "yearmonth" column.
This works as expected:
In [262]: bdata.groupby("yearmonth").apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
Out[262]:
yearmonth
201204 -0.109444
201205 -0.290546
But then I want to sort of "broadcast" these values back to the indices in the original data frame, and save them as constant columns where the dates match.
In [263]: dateGrps = bdata.groupby("yearmonth")
In [264]: dateGrps["MarketReturn"] = dateGrps.apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
/mnt/bos-devrnd04/usr6/home/espears/ws/Research/Projects/python-util/src/util/<ipython-input-264-4a68c8782426> in <module>()
----> 1 dateGrps["MarketReturn"] = dateGrps.apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
TypeError: 'DataFrameGroupBy' object does not support item assignment
I realize this naive assignment should not work. But what is the "right" Pandas idiom for assigning the result of a groupby operation into a new column on the parent dataframe?
In the end, I want a column called "MarketReturn" than will be a repeated constant value for all indices that have matching date with the output of the groupby operation.
One hack to achieve this would be the following:
marketRetsByDate = dateGrps.apply(lambda x: (x["PriceReturn"]*x["MarketCap"]/x["MarketCap"].sum()).sum())
bdata["MarketReturn"] = np.repeat(np.NaN, len(bdata))
for elem in marketRetsByDate.index.values:
bdata["MarketReturn"][bdata["yearmonth"]==elem] = marketRetsByDate.ix[elem]
But this is slow, bad, and unPythonic.
In [97]: df = pandas.DataFrame({'month': np.random.randint(0,11, 100), 'A': np.random.randn(100), 'B': np.random.randn(100)})
In [98]: df.join(df.groupby('month')['A'].sum(), on='month', rsuffix='_r')
Out[98]:
A B month A_r
0 -0.040710 0.182269 0 -0.331816
1 -0.004867 0.642243 1 2.448232
2 -0.162191 0.442338 4 2.045909
3 -0.979875 1.367018 5 -2.736399
4 -1.126198 0.338946 5 -2.736399
5 -0.992209 -1.343258 1 2.448232
6 -1.450310 0.021290 0 -0.331816
7 -0.675345 -1.359915 9 2.722156
While I'm still exploring all of the incredibly smart ways that apply concatenates the pieces it's given, here's another way to add a new column in the parent after a groupby operation.
In [236]: df
Out[236]:
yearmonth return
0 201202 0.922132
1 201202 0.220270
2 201202 0.228856
3 201203 0.277170
4 201203 0.747347
In [237]: def add_mkt_return(grp):
.....: grp['mkt_return'] = grp['return'].sum()
.....: return grp
.....:
In [238]: df.groupby('yearmonth').apply(add_mkt_return)
Out[238]:
yearmonth return mkt_return
0 201202 0.922132 1.371258
1 201202 0.220270 1.371258
2 201202 0.228856 1.371258
3 201203 0.277170 1.024516
4 201203 0.747347 1.024516
As a general rule when using groupby(), if you use the .transform() function pandas will return a table with the same length as your original. When you use other functions like .sum() or .first() then pandas will return a table where each row is a group.
I'm not sure how this works with apply but implementing elaborate lambda functions with transform can be fairly tricky so the strategy that I find most helpful is to create the variables I need, place them in the original dataset and then do my operations there.
If I understand what you're trying to do correctly first you can calculate the total market cap for each group:
bdata['group_MarketCap'] = bdata.groupby('yearmonth')['MarketCap'].transform('sum')
This will add a column called "group_MarketCap" to your original data which would contain the sum of market caps for each group. Then you can calculate the weighted values directly:
bdata['weighted_P'] = bdata['PriceReturn'] * (bdata['MarketCap']/bdata['group_MarketCap'])
And finally you would calculate the weighted average for each group using the same transform function:
bdata['MarketReturn'] = bdata.groupby('yearmonth')['weighted_P'].transform('sum')
I tend to build my variables this way. Sometimes you can pull off putting it all in a single command but that doesn't always work with groupby() because most of the time pandas needs to instantiate the new object to operate on it at the full dataset scale (i.e. you can't add two columns together if one doesn't exist yet).
Hope this helps :)
May I suggest the transform method (instead of aggregate)? If you use it in your original example it should do what you want (the broadcasting).
I did not find a way to make assignment to the original dataframe. So I just store the results from the groups and concatenate them. Then we sort the concatenated dataframe by index to get the original order as the input dataframe. Here is a sample code:
In [10]: df = pd.DataFrame({'month': np.random.randint(0,11, 100), 'A': np.random.randn(100), 'B': np.random.randn(100)})
In [11]: df.head()
Out[11]:
month A B
0 4 -0.029106 -0.904648
1 2 -2.724073 0.492751
2 7 0.732403 0.689530
3 2 0.487685 -1.017337
4 1 1.160858 -0.025232
In [12]: res = []
In [13]: for month, group in df.groupby('month'):
...: new_df = pd.DataFrame({
...: 'A^2+B': group.A ** 2 + group.B,
...: 'A+B^2': group.A + group.B**2
...: })
...: res.append(new_df)
...:
In [14]: res = pd.concat(res).sort_index()
In [15]: res.head()
Out[15]:
A^2+B A+B^2
0 -0.903801 0.789282
1 7.913327 -2.481270
2 1.225944 1.207855
3 -0.779501 1.522660
4 1.322360 1.161495
This method is pretty fast and extensible. You can derive any feature here.
Note: If the dataframe is too large, concat may cause you MMO error.
I'm having some trouble with pandas. I opened a .xlsx file with pandas, but when I try to filter any information, it shows me the error
AttributeError: 'dict' object has no attribute 'head' #(or iloc, or loc, or anything else from DF/pandas)#
So, I did some research and realized that my table turned into a dictionary (why?).
I'm trying to convert this mess into a proper dictionary, so I can convert it into a properly df, because right now, it shows some characteristics from both. I need a df, just it.
Here is the code:
import pandas as pd
df = pd.read_excel('report.xlsx', sheet_name = ["May"])
print(df)
Result: it shows the table plus "[60 rows x 24 columns]"
But when I try to filter or iterate, it shows all dicts possible attibute errors.
Somethings I tried: .from_dict, xls.parse/(df.to_dict).
When I try to convert df to dict properly, it shows
ValueError: If using all scalar values, you must pass an index
I tried this link: [https://stackoverflow.com/questions/17839973/constructing-pandas-dataframe-from-values-in-variables-gives-valueerror-if-usi)][1], but it didn't work. For some reason, it said in one of the errors that I should provide 2-d parameters, that's why I tried to create a new dict and do a sort of 'append', but it didn't work too...
Then I tried all stuff to set an index, but it doesn't let me rename columns because it says .iloc is not an attribute from dict)
I'm new in python, but I never saw a 'pd.read_excel' open a DataFrame as 'dict'. What should I do?
tks!
[1]: Constructing pandas DataFrame from values in variables gives "ValueError: If using all scalar values, you must pass an index"
if its a dict of DataFrames try...
>>> dict_df = {"a":pd.DataFrame([{1:2,3:4},{1:4,4:6}]), "b":pd.DataFrame([{7:9},{1:4}])}
>>> dict_df
{'a': 1 3 4
0 2 4.0 NaN
1 4 NaN 6.0, 'b': 7 1
0 9.0 NaN
1 NaN 4.0}
>>> pd.concat(dict_df.values(),keys=dict_df.keys(), axis=1)
a b
1 3 4 7 1
0 2 4.0 NaN 9.0 NaN
1 4 NaN 6.0 NaN 4.0
I am working on one dataset in which I need to find complete empty columns from the dataset.
example:
A B C D
nan nan nan nan
1 ss nan 3.0
2 bb w2 4.0
nan nan nan nan
Currently, I am using
import pandas as pd
nan_col=[]
for col in df.columns:
if df.loc[df[col].isnull()].empty !=True:
nan_col.append(col)
But this is capturing null values in the specified columns but I need to capture null rows.
expected Answer: row [0,3]
Can anyone suggest me a way to proceed to identify a complete null row in the dataframe.
You can compare if all rows has missing values by DataFrame.isna with DataFrame.all and then get index values by boolean indexing:
L = df.index[df.isna().all(axis=1)].tolist()
#alternative, if huge dataframe slowier
#L = df[df.isna().all(axis=1)].index.tolist()
print (L)
[0, 3]
Or you could use dropna with set and sorted, I get the index after dropping the rows with NaNs and then also get the index of the whole dataframe and use ^ to get the values that aren't in both indexes, then after the I use sorted to sort the list and convert it into a list, like the below:
print(sorted(set(df.index) ^ set(df.dropna(how='all').index)))
If you might have duplicate index, you can do a list comprehension to iterate through the whole df's index, and add the value to the list comprehension if the value isn't in the dropna index, I also use enumerate so that if all indexes are the same (all duplicate index), it would still work, like the below:
idx = df.dropna(how='all').index
print([i for index, i in enumerate(df.index) if index not in idx])
Both codes output:
[0, 3]
I am experimenting with Dask, but I encountered a problem while using apply after grouping.
I have a Dask DataFrame with a large number of rows. Let's consider for example the following
N=10000
df = pd.DataFrame({'col_1':np.random.random(N), 'col_2': np.random.random(N) })
ddf = dd.from_pandas(df, npartitions=8)
I want to bin the values of col_1 and I follow the solution from here
bins = np.linspace(0,1,11)
labels = list(range(len(bins)-1))
ddf2 = ddf.map_partitions(test_f, 'col_1',bins,labels)
where
def test_f(df,col,bins,labels):
return df.assign(bin_num = pd.cut(df[col],bins,labels=labels))
and this works as I expect it to.
Now I want to take the median value in each bin (taken from here)
median = ddf2.groupby('bin_num')['col_1'].apply(pd.Series.median).compute()
Having 10 bins, I expect median to have 10 rows, but it actually has 80. The dataframe has 8 partitions so I guess that somehow the apply is working on each one individually.
However, If I want the mean and use mean
median = ddf2.groupby('bin_num')['col_1'].mean().compute()
it works and the output has 10 rows.
The question is then: what am I doing wrong that is preventing apply from operating as mean?
Maybe this warning is the key (Dask doc: SeriesGroupBy.apply) :
Pandas’ groupby-apply can be used to to apply arbitrary functions, including aggregations that result in one row per group. Dask’s groupby-apply will apply func once to each partition-group pair, so when func is a reduction you’ll end up with one row per partition-group pair. To apply a custom aggregation with Dask, use dask.dataframe.groupby.Aggregation.
You are right! I was able to reproduce your problem on Dask 2.11.0. The good news is that there's a solution! It appears that the Dask groupby problem is specifically with the category type (pandas.core.dtypes.dtypes.CategoricalDtype). By casting the category column to another column type (float, int, str), then the groupby will work correctly.
Here's your code that I copied:
import dask.dataframe as dd
import pandas as pd
import numpy as np
def test_f(df, col, bins, labels):
return df.assign(bin_num=pd.cut(df[col], bins, labels=labels))
N = 10000
df = pd.DataFrame({'col_1': np.random.random(N), 'col_2': np.random.random(N)})
ddf = dd.from_pandas(df, npartitions=8)
bins = np.linspace(0,1,11)
labels = list(range(len(bins)-1))
ddf2 = ddf.map_partitions(test_f, 'col_1', bins, labels)
print(ddf2.groupby('bin_num')['col_1'].apply(pd.Series.median).compute())
which prints out the problem you mentioned
bin_num
0 NaN
1 NaN
2 NaN
3 NaN
4 NaN
...
5 0.550844
6 0.651036
7 0.751220
8 NaN
9 NaN
Name: col_1, Length: 80, dtype: float64
Here's my solution:
ddf3 = ddf2.copy()
ddf3["bin_num"] = ddf3["bin_num"].astype("int")
print(ddf3.groupby('bin_num')['col_1'].apply(pd.Series.median).compute())
which printed:
bin_num
9 0.951369
2 0.249150
1 0.149563
0 0.049897
3 0.347906
8 0.847819
4 0.449029
5 0.550608
6 0.652778
7 0.749922
Name: col_1, dtype: float64
#MRocklin or #TomAugspurger
Would you be able to create a fix for this in a new release? I think there is sufficient reproducible code here. Thanks for all your hard work. I love Dask and use it every day ;)
How can I find the row for which the value of a specific column is maximal?
df.max() will give me the maximal value for each column, I don't know how to get the corresponding row.
Use the pandas idxmax function. It's straightforward:
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].idxmax()
3
>>> df['B'].idxmax()
4
>>> df['C'].idxmax()
1
Alternatively you could also use numpy.argmax, such as numpy.argmax(df['A']) -- it provides the same thing, and appears at least as fast as idxmax in cursory observations.
idxmax() returns indices labels, not integers.
Example': if you have string values as your index labels, like rows 'a' through 'e', you might want to know that the max occurs in row 4 (not row 'd').
if you want the integer position of that label within the Index you have to get it manually (which can be tricky now that duplicate row labels are allowed).
HISTORICAL NOTES:
idxmax() used to be called argmax() prior to 0.11
argmax was deprecated prior to 1.0.0 and removed entirely in 1.0.0
back as of Pandas 0.16, argmax used to exist and perform the same function (though appeared to run more slowly than idxmax).
argmax function returned the integer position within the index of the row location of the maximum element.
pandas moved to using row labels instead of integer indices. Positional integer indices used to be very common, more common than labels, especially in applications where duplicate row labels are common.
For example, consider this toy DataFrame with a duplicate row label:
In [19]: dfrm
Out[19]:
A B C
a 0.143693 0.653810 0.586007
b 0.623582 0.312903 0.919076
c 0.165438 0.889809 0.000967
d 0.308245 0.787776 0.571195
e 0.870068 0.935626 0.606911
f 0.037602 0.855193 0.728495
g 0.605366 0.338105 0.696460
h 0.000000 0.090814 0.963927
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
In [20]: dfrm['A'].idxmax()
Out[20]: 'i'
In [21]: dfrm.iloc[dfrm['A'].idxmax()] # .ix instead of .iloc in older versions of pandas
Out[21]:
A B C
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
So here a naive use of idxmax is not sufficient, whereas the old form of argmax would correctly provide the positional location of the max row (in this case, position 9).
This is exactly one of those nasty kinds of bug-prone behaviors in dynamically typed languages that makes this sort of thing so unfortunate, and worth beating a dead horse over. If you are writing systems code and your system suddenly gets used on some data sets that are not cleaned properly before being joined, it's very easy to end up with duplicate row labels, especially string labels like a CUSIP or SEDOL identifier for financial assets. You can't easily use the type system to help you out, and you may not be able to enforce uniqueness on the index without running into unexpectedly missing data.
So you're left with hoping that your unit tests covered everything (they didn't, or more likely no one wrote any tests) -- otherwise (most likely) you're just left waiting to see if you happen to smack into this error at runtime, in which case you probably have to go drop many hours worth of work from the database you were outputting results to, bang your head against the wall in IPython trying to manually reproduce the problem, finally figuring out that it's because idxmax can only report the label of the max row, and then being disappointed that no standard function automatically gets the positions of the max row for you, writing a buggy implementation yourself, editing the code, and praying you don't run into the problem again.
You might also try idxmax:
In [5]: df = pandas.DataFrame(np.random.randn(10,3),columns=['A','B','C'])
In [6]: df
Out[6]:
A B C
0 2.001289 0.482561 1.579985
1 -0.991646 -0.387835 1.320236
2 0.143826 -1.096889 1.486508
3 -0.193056 -0.499020 1.536540
4 -2.083647 -3.074591 0.175772
5 -0.186138 -1.949731 0.287432
6 -0.480790 -1.771560 -0.930234
7 0.227383 -0.278253 2.102004
8 -0.002592 1.434192 -1.624915
9 0.404911 -2.167599 -0.452900
In [7]: df.idxmax()
Out[7]:
A 0
B 8
C 7
e.g.
In [8]: df.loc[df['A'].idxmax()]
Out[8]:
A 2.001289
B 0.482561
C 1.579985
Both above answers would only return one index if there are multiple rows that take the maximum value. If you want all the rows, there does not seem to have a function.
But it is not hard to do. Below is an example for Series; the same can be done for DataFrame:
In [1]: from pandas import Series, DataFrame
In [2]: s=Series([2,4,4,3],index=['a','b','c','d'])
In [3]: s.idxmax()
Out[3]: 'b'
In [4]: s[s==s.max()]
Out[4]:
b 4
c 4
dtype: int64
df.iloc[df['columnX'].argmax()]
argmax() would provide the index corresponding to the max value for the columnX. iloc can be used to get the row of the DataFrame df for this index.
A more compact and readable solution using query() is like this:
import pandas as pd
df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
print(df)
# find row with maximum A
df.query('A == A.max()')
It also returns a DataFrame instead of Series, which would be handy for some use cases.
Very simple: we have df as below and we want to print a row with max value in C:
A B C
x 1 4
y 2 10
z 5 9
In:
df.loc[df['C'] == df['C'].max()] # condition check
Out:
A B C
y 2 10
If you want the entire row instead of just the id, you can use df.nlargest and pass in how many 'top' rows you want and you can also pass in for which column/columns you want it for.
df.nlargest(2,['A'])
will give you the rows corresponding to the top 2 values of A.
use df.nsmallest for min values.
The direct ".argmax()" solution does not work for me.
The previous example provided by #ely
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].argmax()
3
>>> df['B'].argmax()
4
>>> df['C'].argmax()
1
returns the following message :
FutureWarning: 'argmax' is deprecated, use 'idxmax' instead. The behavior of 'argmax'
will be corrected to return the positional maximum in the future.
Use 'series.values.argmax' to get the position of the maximum now.
So that my solution is :
df['A'].values.argmax()
mx.iloc[0].idxmax()
This one line of code will give you how to find the maximum value from a row in dataframe, here mx is the dataframe and iloc[0] indicates the 0th index.
Considering this dataframe
[In]: df = pd.DataFrame(np.random.randn(4,3),columns=['A','B','C'])
[Out]:
A B C
0 -0.253233 0.226313 1.223688
1 0.472606 1.017674 1.520032
2 1.454875 1.066637 0.381890
3 -0.054181 0.234305 -0.557915
Assuming one want to know the rows where column "C" is max, the following will do the work
[In]: df[df['C']==df['C'].max()])
[Out]:
A B C
1 0.472606 1.017674 1.520032
The idmax of the DataFrame returns the label index of the row with the maximum value and the behavior of argmax depends on version of pandas (right now it returns a warning). If you want to use the positional index, you can do the following:
max_row = df['A'].values.argmax()
or
import numpy as np
max_row = np.argmax(df['A'].values)
Note that if you use np.argmax(df['A']) behaves the same as df['A'].argmax().
Use:
data.iloc[data['A'].idxmax()]
data['A'].idxmax() -finds max value location in terms of row
data.iloc() - returns the row
If there are ties in the maximum values, then idxmax returns the index of only the first max value. For example, in the following DataFrame:
A B C
0 1 0 1
1 0 0 1
2 0 0 0
3 0 1 1
4 1 0 0
idxmax returns
A 0
B 3
C 0
dtype: int64
Now, if we want all indices corresponding to max values, then we could use max + eq to create a boolean DataFrame, then use it on df.index to filter out indexes:
out = df.eq(df.max()).apply(lambda x: df.index[x].tolist())
Output:
A [0, 4]
B [3]
C [0, 1, 3]
dtype: object
what worked for me is:
df[df['colX'] == df['colX'].max()
You then get the row in your df with the maximum value of colX.
Then if you just want the index you can add .index at the end of the query.