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why is pd.crosstab not giving the expected output in python pandas?

I have a 2dataframes, which I am calling as df1 and df2.
df1 has columns like KPI and context and it looks like this.
KPI Context
0 Does the company have a policy in place to man... Anti-Bribery Policy\nBroadridge does not toler...
1 Does the company have a supplier code of conduct? Vendor Code of Conduct Our vendors play an imp...
2 Does the company have a grievance/complaint ha... If you ever have a question or wish to report ...
3 Does the company have a human rights policy ? Human Rights Statement of Commitment Broadridg...
4 Does the company have a policies consistent wi... Anti-Bribery Policy\nBroadridge does not toler...
df2 has a single column 'keyword'
df2:
Keyword
0 1.5 degree
1 1.5°
2 2 degree
3 2°
4 accident
I wanted to create another dataframe out of these two dataframe wherein if a particular value from 'Keyword' column of df2 is present in the 'Context' of df1 then simply write the count of it.
for which I have used pd.crosstab() however I suspect that its not giving me the expected output.
here's what I have tried so far.
new_df = df1.explode('Context')
new_df1 = df2.explode('Keyword')
new_df = pd.crosstab(new_df['KPI'], new_df1['Keyword'], values=new_df['Context'], aggfunc='count').reset_index().rename_axis(columns=None)
print(new_df.head())
the new_df looks like this.
KPI 1.5 degree 1.5° \
0 Does the Supplier code of conduct cover one or... NaN NaN
1 Does the companies have sites/operations locat... NaN NaN
2 Does the company have a due diligence process ... NaN NaN
3 Does the company have a grievance/complaint ha... NaN NaN
4 Does the company have a grievance/complaint ha... NaN NaN
2 degree 2° accident
0 NaN NaN NaN
1 NaN NaN NaN
2 NaN NaN NaN
3 1.0 NaN NaN
4 NaN NaN NaN
The expected output which I want is something like this.
0 KPI 1.5 degree 1.5° 2 degree 2° accident
1 Does the company have a policy in place to man 44 2 3 5 9
what exactly am I missing? please let me know, thanks!

There is multiple problems - first explode working with splitted values, not with strings. Then for extract Keyword from Context need Series.str.findall and for crosstab use columns in same DataFrame, not 2 different:
import re
pat = '|'.join(r"\b{}\b".format(re.escape(x)) for x in df2['Keyword'])
df1['new'] = df1['Context'].str.findall(pat, flags=re.I)
new_df = df1.explode('new')
out = pd.crosstab(new_df['KPI'], new_df['new'])

Can't fill nan values in pandas even with inplace flag

I have a pandas dataframe containing NaN values for some column.
I'm trying to fill them with a default value (30), but it doesn't work.
Original dataframe:
type avg_speed
0 CAR 32.0
1 CAR NaN
2 CAR NaN
3 BIKE 16.2
4 CAR 28.5
5 SCOOTER 29.7
6 CAR 30.7
7 CAR NaN
8 BIKE NaN
9 BIKE 35.1
...
Desired result:
type avg_speed
0 CAR 32.0
1 CAR 30
2 CAR 30
3 BIKE 16.2
4 CAR 28.5
5 SCOOTER 29.7
6 CAR 30.7
7 CAR 30
8 BIKE 30
9 BIKE 35.1
My code:
def fill_with_default(pandas_df, column_name, default_value):
print(f"Total count: {pandas_df.count()}")
print(f"Count of Nan BEFORE: {pandas_df[column_name].isna().sum()}")
pandas_df[column_name].fillna(default_value, inplace=True)
print(f"Count of Nan AFTER: {pandas_df[column_name].isna().sum()}")
return pandas_df
df = fill_with_default(df, "avg_speed", 30)
Output:
Total count: 105018
Count of Nan BEFORE: 49514
Count of Nan AFTER: 49514
The chain of dataframe transformations and list of columns are too long, so it's difficult to show all steps (join with another dataframe, drop useless columns, add usefull columns, join with other dataframes, filter etc.)
I've tried other options but they also don't work:
#pandas_df.fillna({column_name: default_value}, inplace=True)
#pandas_df.loc[pandas_df[column_name].isnull(),column_name] = default_value
...
Type of column before applying "fillna" is fload64, the same as default_value
Therefore, my question is: what could be the potential reasons of this problem?
What kind of transformation can lead to this problem? Because this is the method that works for another similar data frame. The only difference between them lies in the chain of transformations.
BTW, there is a system log at this place:
/home/hadoop/.local/lib/python3.6/site-
packages/pandas/core/generic.py:6287: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame
See the caveats in the documentation:
http://pandas.pydata.org/pandas-
docs/stable/user_guide/indexing.html#returning-a-view-versus-a-
copy
self._update_inplace(new_data)

How to iterate over columns and check condition by group

I have data for many countries over a period of time (2001-2003). It looks something like this:
index
year
country
inflation
GDP
1
2001
AFG
nan
48
2
2002
AFG
nan
49
3
2003
AFG
nan
50
4
2001
CHI
3.0
nan
5
2002
CHI
5.0
nan
6
2003
CHI
7.0
nan
7
2001
USA
nan
220
8
2002
USA
4.0
250
9
2003
USA
2.5
280
I want to drop countries in case there is no data (i.e. values are missing for all years) for any given variable.
In the example table above, I want to drop AFG (because it misses all values for inflation) and CHI (GDP missing). I don't want to drop observation #7 just because one year is missing.
What's the best way to do that?

This should work by filtering all values that have nan in one of (inflation, GDP):
(
df.groupby(['country'])
.filter(lambda x: not x['inflation'].isnull().all() and not x['GDP'].isnull().all())
)
Note, if you have more than two columns you can work on a more general version of this:
df.groupby(['country']).filter(lambda x: not x.isnull().all().any())
If you want this to work with a specific range of year instead of all columns, you can set up a mask and change the code a bit:
mask = (df['year'] >= 2002) & (df['year'] <= 2003) # mask of years
grp = df.groupby(['country']).filter(lambda x: not x[mask].isnull().all().any())

You can also try this:
# check where the sum is equal to 0 - means no values in the column for a specific country
group_by = df.groupby(['country']).agg({'inflation':sum, 'GDP':sum}).reset_index()
# extract only countries with information on both columns
indexes = group_by[ (group_by['GDP'] != 0) & ( group_by['inflation'] != 0) ].index
final_countries = list(group_by.loc[ group_by.index.isin(indexes), : ]['country'])
# keep the rows contains the countries
df = df.drop(df[~df.country.isin(final_countries)].index)

You could reshape the data frame from long to wide, drop nulls, and then convert back to wide.
To convert from long to wide, you can use pivot functions. See this question too.
Here's code for dropping nulls, after its reshaped:
df.dropna(axis=0, how= 'any', thresh=None, subset=None, inplace=True) # Delete rows, where any value is null
To convert back to long, you can use pd.melt.

Can I cluster these records without having to run these loops for every record?

So I want to cluster the records in this table to find which records are 'similar' (i.e. have enough in common). An example of the table is as follows:
author beginpage endpage volume publication year id_old id_new
0 NaN 495 497 NaN 1975 1 1
1 NaN 306 317 14 1997 2 2
2 lowry 265 275 193 1951 3 3
3 smith p k 76 85 150 1985 4 4
4 NaN 248 254 NaN 1976 5 5
5 hamill p 85 100 391 1981 6 6
6 NaN 1513 1523 7 1979 7 7
7 b oregan 737 740 353 1991 8 8
8 NaN 503 517 98 1975 9 9
9 de wijs 503 517 98 1975 10 10
In this small table, the last row should get 'new_id' equal to 9, to show that these two records are similar.
To make this happen I wrote the code below, which works fine for a small number of records. However, I want to use my code for a table with 15000 records. And of course, if you do the maths, with this code this is going to take way too long.
Anyone who could help me make this code more efficient? Thanks in advance!
My code, where 'dfhead' is the table with the records:
for r in range(0,len(dfhead)):
for o_r in range(r+1,len(dfhead)):
if ((dfhead.loc[r,c] == dfhead.loc[o_r,c]).sum() >= 3) :
if (dfhead.loc[o_r,['id_new']] > dfhead.loc[r,['id_new']]).sum() ==1:
dfhead.loc[o_r,['id_new']] = dfhead.loc[r,['id_new']]

If you are only trying to detect whole equalities between "beginpage", "endpage","volume", "publication", "year", you should try to work on duplicates. I'm not sure about this as your code is still a mistery for me.
Something like this might work (your column "id" needs to be named "id_old" at first in the dataframe though):
cols = ["beginpage", "endpage","volume", "publication", "year"]
#isolate duplicated rows
duplicated = df[df.duplicated(cols, keep=False)]
#find the minimum key to keep
temp = duplicated.groupby(cols, as_index=False)['index'].min()
temp.rename({'id_old':'id_new'}, inplace=True, axis=1)
#import the "minimum key" to duplicated by merging the dataframes
duplicated = duplicated.merge(temp, on=cols, how="left")
#gather the "un-duplicated" rows
unduplicated = df[~df.duplicated(cols, keep=False)]
#concatenate both datasets and reset the index
new_df = unduplicated.append(duplicated)
new_df.reset_index(drop=True, inplace=True)
#where "id_new" is empty, then the data comes from "unduplicated"
#and you could fill the datas from id_old
ix = new_df[new_df.id_new.isnull()].index
new_df.loc[ix, 'id_new'] = new_df.loc[ix, 'id_old']

Pandas - Merge rows on column A, taking first values from each column B, C etc

I have a dataframe, with recordings of statistics in multiple columns.
I have a list of the column names: stat_columns = ['Height', 'Speed'].
I want to combine the data to get one row per id.
The data comes sorted with the newest records on the top. I want the most recent data, so I must use the first value of each column, by id.
My dataframe looks like this:
Index id Height Speed
0 100007 8.3
1 100007 54
2 100007 8.6
3 100007 52
4 100035 39
5 100014 44
6 100035 5.6
And I want it to look like this:
Index id Height Speed
0 100007 54 8.3
1 100014 44
2 100035 39 5.6
I have tried a simple groupby myself:
df_stats = df_path.groupby(['id'], as_index=False).first()
But this seems to only give me a row with the first statistic found.

For me your solution working, maybe is necessary replace empty values to NaNs:
df_stats = df_path.replace('',np.nan).groupby('id', as_index=False).first()
print (df_stats)
id Index Height Speed
0 100007 0 54.0 8.3
1 100014 5 44.0 NaN
2 100035 4 39.0 5.6