I have two dataframes both of which have the same basic schema. (4 date fields, a couple of string fields, and 4-5 float fields). Call them df1 and df2.
What I want to do is basically get a "diff" of the two - where I get back all rows that are not shared between the two dataframes (not in the set intersection). Note, the two dataframes need not be the same length.
I tried using pandas.merge(how='outer') but I was not sure what column to pass in as the 'key' as there really isn't one and the various combinations I tried were not working. It is possible that df1 or df2 has two (or more) rows that are identical.
What is a good way to do this in pandas/Python?
Try this:
diff_df = pd.merge(df1, df2, how='outer', indicator='Exist')
diff_df = diff_df.loc[diff_df['Exist'] != 'both']
You will have a dataframe of all rows that don't exist on both df1 and df2.
IIUC:
You can use pd.Index.symmetric_difference
pd.concat([df1, df2]).loc[
df1.index.symmetric_difference(df2.index)
]
You can use this function, the output is an ordered dict of 6 dataframes which you can write to excel for further analysis.
'df1' and 'df2' refers to your input dataframes.
'uid' refers to the column or combination of columns that make up the unique key. (i.e. 'Fruits')
'dedupe' (default=True) drops duplicates in df1 and df2. (refer to Step 4 in comments)
'labels' (default = ('df1','df2')) allows you to name the input dataframes. If a unique key exists in both dataframes, but have
different values in one or more columns, it is usually important to know these rows, put them one on top of the other and label the row with the name so we know to which dataframe does it belong to.
'drop' can take a list of columns to be excluded from the consideration when considering the difference
Here goes:
df1 = pd.DataFrame([['apple', '1'], ['banana', 2], ['coconut',3]], columns=['Fruits','Quantity'])
df2 = pd.DataFrame([['apple', '1'], ['banana', 3], ['durian',4]], columns=['Fruits','Quantity'])
dict1 = diff_func(df1, df2, 'Fruits')
In [10]: dict1['df1_only']:
Out[10]:
Fruits Quantity
1 coconut 3
In [11]: dict1['df2_only']:
Out[11]:
Fruits Quantity
3 durian 4
In [12]: dict1['Diff']:
Out[12]:
Fruits Quantity df1 or df2
0 banana 2 df1
1 banana 3 df2
In [13]: dict1['Merge']:
Out[13]:
Fruits Quantity
0 apple 1
Here is the code:
import pandas as pd
from collections import OrderedDict as od
def diff_func(df1, df2, uid, dedupe=True, labels=('df1', 'df2'), drop=[]):
dict_df = {labels[0]: df1, labels[1]: df2}
col1 = df1.columns.values.tolist()
col2 = df2.columns.values.tolist()
# There could be columns known to be different, hence allow user to pass this as a list to be dropped.
if drop:
print ('Ignoring columns {} in comparison.'.format(', '.join(drop)))
col1 = list(filter(lambda x: x not in drop, col1))
col2 = list(filter(lambda x: x not in drop, col2))
df1 = df1[col1]
df2 = df2[col2]
# Step 1 - Check if no. of columns are the same:
len_lr = len(col1), len(col2)
assert len_lr[0]==len_lr[1], \
'Cannot compare frames with different number of columns: {}.'.format(len_lr)
# Step 2a - Check if the set of column headers are the same
# (order doesnt matter)
assert set(col1)==set(col2), \
'Left column headers are different from right column headers.' \
+'\n Left orphans: {}'.format(list(set(col1)-set(col2))) \
+'\n Right orphans: {}'.format(list(set(col2)-set(col1)))
# Step 2b - Check if the column headers are in the same order
if col1 != col2:
print ('[Note] Reordering right Dataframe...')
df2 = df2[col1]
# Step 3 - Check datatype are the same [Order is important]
if set((df1.dtypes == df2.dtypes).tolist()) - {True}:
print ('dtypes are not the same.')
df_dtypes = pd.DataFrame({labels[0]:df1.dtypes,labels[1]:df2.dtypes,'Diff':(df1.dtypes == df2.dtypes)})
df_dtypes = df_dtypes[df_dtypes['Diff']==False][[labels[0],labels[1],'Diff']]
print (df_dtypes)
else:
print ('DataType check: Passed')
# Step 4 - Check for duplicate rows
if dedupe:
for key, df in dict_df.items():
if df.shape[0] != df.drop_duplicates().shape[0]:
print(key + ': Duplicates exists, they will be dropped.')
dict_df[key] = df.drop_duplicates()
# Step 5 - Check for duplicate uids.
if type(uid)==str or type(uid)==list:
print ('Uniqueness check: {}'.format(uid))
for key, df in dict_df.items():
count_uid = df.shape[0]
count_uid_unique = df[uid].drop_duplicates().shape[0]
var = [0,1][count_uid_unique == df.shape[0]] #<-- Round off to the nearest integer if it is 100%
pct = round(100*count_uid_unique/df.shape[0], var)
print ('{}: {} out of {} are unique ({}%).'.format(key, count_uid_unique, count_uid, pct))
# Checks complete, begin merge. '''Remenber to dedupe, provide labels for common_no_match'''
dict_result = od()
df_merge = pd.merge(df1, df2, on=col1, how='inner')
if not df_merge.shape[0]:
print ('Error: Merged DataFrame is empty.')
else:
dict_result[labels[0]] = df1
dict_result[labels[1]] = df2
dict_result['Merge'] = df_merge
if type(uid)==str:
uid = [uid]
if type(uid)==list:
df1_only = df1.append(df_merge).reset_index(drop=True)
df1_only['Duplicated']=df1_only.duplicated(keep=False) #keep=False, marks all duplicates as True
df1_only = df1_only[df1_only['Duplicated']==False]
df2_only = df2.append(df_merge).reset_index(drop=True)
df2_only['Duplicated']=df2_only.duplicated(keep=False)
df2_only = df2_only[df2_only['Duplicated']==False]
label = labels[0]+' or '+labels[1]
df_lc = df1_only.copy()
df_lc[label] = labels[0]
df_rc = df2_only.copy()
df_rc[label] = labels[1]
df_c = df_lc.append(df_rc).reset_index(drop=True)
df_c['Duplicated'] = df_c.duplicated(subset=uid, keep=False)
df_c1 = df_c[df_c['Duplicated']==True]
df_c1 = df_c1.drop('Duplicated', axis=1)
df_uc = df_c[df_c['Duplicated']==False]
df_uc_left = df_uc[df_uc[label]==labels[0]]
df_uc_right = df_uc[df_uc[label]==labels[1]]
dict_result[labels[0]+'_only'] = df_uc_left.drop(['Duplicated', label], axis=1)
dict_result[labels[1]+'_only'] = df_uc_right.drop(['Duplicated', label], axis=1)
dict_result['Diff'] = df_c1.sort_values(uid).reset_index(drop=True)
return dict_result
Set df2.columns = df1.columns
Now, set every column as the index: df1 = df1.set_index(df1.columns.tolist()), and similarly for df2.
You can now do df1.index.difference(df2.index), and df2.index.difference(df1.index), and the two results are your distinct columns.
with
left_df.merge(df,left_on=left_df.columns.tolist(),right_on=df.columns.tolist(),how='outer')
you can get the outer join result.
Similarly, you can get the inner join result.Then make a diff that would be what you want.
Related
I have a pandas DataFrame 'df' with x rows, and another pandas DataFrame 'df2' with y rows
(x < y). I want to return the indexes of where the values of df['Farm'] equals the value of df2['Fields'], in order to add respective 'Manager' to df.
the code I have is as follows:
data2 = [['field1', 'Paul G'] , ['field2', 'Mark R'], ['field3', 'Roy Jr']]
data = [['field1'] , ['field2']]
columns = ['Field']
columns2 = ['Field', 'Manager']
df = pd.DataFrame(data, columns=columns)
df2 = pd.DataFrame(data2, columns=columns2)
farmNames = df['Farm']
exists = farmNames.reset_index(drop=True) == df1['Field'].reset_index(drop=True)
This returns the error message:
ValueError: Can only compare identically-labeled Series objects
Does anyone know how to fix this?
As #NickODell mentioned, you could use a merge, basically a left join. See below code.
df_new = pd.merge(df, df2, on = 'Field', how = 'left')
print(df_new)
Output:
Field Manager
0 field1 Paul G
1 field2 Mark R
This loop is intended to match subjects in df2 to subjects in df1 with a 1:4 ratio. The key here is randomly selecting subjects while avoiding redundancy. No subject should be matched twice. df1 has a few thousand subjects, whereas df2 has over one million. Every subject in df1 will be matched to four subjects in df2, those who aren't matched will be left out. Does anyone have ideas for improving its efficiency? An approach that also conserves RAM would be ideal. Thanks.
for x in range(4): # 1:4 matching
for index, row in df1.iterrows():
temp = df2.loc[(df2['matched'] != 1) & (df2['race_ethnicity'] == row['race_ethnicity']) & (df2['age'] == row['age']) & ((df2['date1'] > row['date2']) | (df2['date1'].isna()))]
a = temp.sample()
a['matched_subject'] = row['subject_id']
a['matched'] = '1'
a['possible_matches'] = len(temp)
It can be simplified to this, but I'd prefer to continue using the 'possible_matches' row for diagnostics.
for x in range(4): # 4 because 1:4 matching
for index, row in df1.iterrows():
a = df2.loc[(df2['matched_subject']=='') & (df2['race_ethnicity'] == row['race_ethnicity']) & (df2['age'] == row['age']) & ((df2['date1'] > row['date2']) | (df2['date1'] == 0))].sample()
a['matched_subject'] = row['pid']
Clarifications: All rows are unique in both DataFrames, representing a list of subjects that will be compared in survival analysis. Date1 is the datetime of outcome variable event, is present for a fraction of subjects in both DataFrames. Date 2 is the datetime of independent variable event, present for all df1 subjects and no df2 subjects. Sample inputs include:
subject_id (numeric)
race_ethnicity (str, 3 categories)
age (numeric)
Matched (binary, indicates if a df2 subject has been matched to one in df1)
Matched_subject (numeric, the subject_id of the matched subject)
date1 (datetime. outcome variable for survival models, present for some and not others in both dataframes)
date2 (datetime for event that is independent variable. Everyone in df1 has a date2, nobody in df2 has a date2. A df1 subject's date2 is their index date for survival analysis, and also serves as index date for their matched subjects who have no date2 variable)
We want to match four subjects in df2 for every one in df1. Date2 in df1 will be the index date for matched subjects in df2, hence we have a condition to ensure that date1 (outcome event) in df2 doesn't occur before date2 in df1 (index event)
Here is an example of two df2 subjects matched to df1. Also with an unmatched df1 and df2 subject. The real df2 is large enough to match all df1 subjects with great excess.
df1:
subject_id
race_ethnicity
age
date1
date2
matched
matched_subject
possible_matches
3a3r796e
Non-Hispanic white
55
(can be present in df1 or df2. if present, must be after date 2 in df1)
2012-01-01
1
3a3r796e matching based on df1, therefore these are the same
only important for df2. not important for analysis, just a diagnostic value
1234abcd
Non-Hispanic black
58
2017-01-01
2016-01-01
0
df2:
subject_id
race_ethnicity
age
date1
date2
matched
matched_subject
possible_matches
5c69a756
Non-Hispanic white
55
2015-01-01
(cannot be present in df2, by definition)
1
3a3r796e
571
7as89f75
Non-Hispanic white
55
1
3a3r796e
571
6376asef
Hispanic
42
2010-01-01
0
First let's define some helper functions:
def generate_data(df1_len, df2_len, seed=42):
"""Generate random data to help test different algorithms"""
np.random.seed(seed)
d2 = np.random.randint(0, 3000, size=df1_len)
df1 = pd.DataFrame({
'subject_id': np.arange(df1_len),
'race_ethnicity': np.random.choice(list('ABC'), df1_len),
'age': np.random.randint(18, 100, df1_len),
'date2': pd.Timestamp('2000-01-01') + pd.to_timedelta(d2, unit='D')
})
d1 = np.random.randint(0, 3000, size=int(df2_len * np.random.rand()))
d1 = np.hstack([d2, np.repeat(np.nan, df2_len - len(d1))])
df2 = pd.DataFrame({
'subject_id': np.arange(df2_len),
'race_ethnicity': np.random.choice(list('ABC'), df2_len),
'age': np.random.randint(18, 100, df2_len),
'date1': pd.Timestamp('2000-01-01') + pd.to_timedelta(d1, unit='D')
})
return df1, df2
def verify(df1, df2):
"""Verify that df1 and df2 are matched according to predefined rules"""
tmp = df1.merge(df2, how='left', left_on='subject_id', right_on='matched_subject', suffixes=('_1', "_2"))
assert (tmp['race_ethnicity_1'] == tmp['race_ethnicity_2']).all(), 'race_ethnicity does not match'
assert (tmp['age_1'] == tmp['age_2']).all(), 'age does not match'
assert ((tmp['date1'] > tmp['date2']) | tmp['date1'].isna()).all(), 'date1 must be NaT or grater than date2'
assert tmp.groupby('matched_subject').size().eq(4).all(), 'Invalid match ratio'
print('All is good')
The original solution
Allow me to make some changes in the interest of clarity. This version runs in
~28 seconds on my Mac:
df1, df2 = generate_data(500, 100_000)
df2['matched'] = False
df2['matched_subject'] = None
df2['possible_matches'] = None
for x in range(4): # 1:4 matching
for index, row in df1.iterrows():
cond = (
(df2['matched'] != 1) &
(df2['race_ethnicity'] == row['race_ethnicity']) &
(df2['age'] == row['age']) &
((df2['date1'] > row['date2']) | df2['date1'].isna())
)
temp = df2.loc[cond]
if temp.empty:
continue
idx = temp.sample().index
df2.loc[idx, 'matched_subject'] = row['subject_id']
df2.loc[idx, 'matched'] = True
df2.loc[idx, 'possible_matches'] = len(temp)
An improved version
By taking out the outer loop (for _ in range(4)), you can improve performance
almost 4 times. The code below executed in 7s:
df1, df2 = generate_data(5000, 1_000_000)
df2['matched'] = False
df2['matched_subject'] = None
df2['possible_matches'] = None
for index, row in df1.iterrows():
cond = (
(df2['matched'] != 1) &
(df2['race_ethnicity'] == row['race_ethnicity']) &
(df2['age'] == row['age']) &
((df2['date1'] > row['date2']) | df2['date1'].isna())
)
temp = df2.loc[cond]
if temp.empty:
continue
idx = temp.sample(4).index
df2.loc[idx, 'matched_subject'] = row['subject_id']
df2.loc[idx, 'matched'] = True
df2.loc[idx, 'possible_matches'] = len(temp)
A further improved version
Taking the idea that working on multiple rows at once is faster than doing it
one at a time, we can loop based on group of rows with similar characteristics
rather than looping with individual rows. This code runs in 600ms or ~46x faster
than the original version:
df1, df2 = generate_data(500, 100_000)
# Shuffle df2 so the matches will be random
df2 = df2.sample(frac=1)
# A dictionary to hold the result. Its keys are the indexes in df2 and its
# values are the indexes of df1
matches = {}
# We loop by group instead of individual row
grouped1 = df1.groupby(['race_ethnicity', 'age', 'date2'])
grouped2 = df2.groupby(['race_ethnicity', 'age'])
for (race_ethnicity, age, date2), subset1 in grouped1:
# Get all rows from df2 that have the same `race_ethnicity` and `age`
subset2 = grouped2.get_group((race_ethnicity, age))
# pd.Series is slow. Switch to np.array for speed
index2 = subset2.index.to_numpy()
date1 = subset2['date1'].to_numpy()
# Since all rows in subset1 and subset2 have already had the same
# `race_ethnicity` and `age`, we only need to filter for two things:
# 1. The relationship between `date1` and `date2`; and
# 2. That the row in `df2` has NOT been matched before
cond = (
(np.isnan(date1) | (date1 > date2))
& np.isin(index2, list(matches.keys()), invert=True)
)
# The match ratio
index1 = np.repeat(subset1.index.to_numpy(), 4)
# There is no way to know in advance how many rows in `subset2` will meet
# the matching criteria:
# * Ideally: cond.sum() == len(index1), ie. 4 rows in `subset2` for every
# row in `subset1`
# * If there are more matches than we need: we will take the first `4 *
# len(subset1)` rows
# * If there are not enough matches: eg. 6 rows in `subset2` for 2 rows in
# `subset1`, some rows in `subset1` will have to accept < 4 matches
n = min(cond.sum(), len(index1))
matches.update({
key: value for key, value in zip(index2[cond][:n], index1[:n])
})
tmp = pd.DataFrame({
'index2': matches.keys(),
'index1': matches.values()
})
df2 = (
df2.merge(tmp, how='left', left_index=True, right_on='index2')
.merge(df1['subject_id'].to_frame('matched_subject'), how='left', left_on='index1', right_index=True)
.drop(columns=['index1', 'index2'])
)
You can verify the solution:
verify(df1, df2)
# Output: All is good
If I need to map one value between two dataframes, and get 'FD' value from row where Round = 1 an Id is 262:
df1 = pd.DataFrame({'Round':1,'ID':262,'FD':30,
'Round':2,'ID':262,'FD':20}, index=[0])
df2 = pd.DataFrame({'Round':1, 'Opponent':262,
'Round':2, 'Opponent':262},index=[0])
I have tried to map with:
df2['P_GS_by_FD'] = f2['Opponent'].map(df1.set_index('ID')['FD'])
df2 Expected output:
Round Opponent P_GS_by_FD
1 262 30
I would use drop_duplicates
this would select 'Round 1' rows
df1.drop_duplicates('Id', keep='first')
df2['P_GS_by_FD'] = df2['Opponent'].map(df1.drop_duplicates('Id', keep='first').set_index('Id')['FD'])
(I think your example df1, df2 would make only one row instead of two)
Then we need create the round in df2 as well
df2['Round'] = df.groupby('Opponent').cumcount()+1
yourdf = df2.merge(df1.rename(columns={'Id' : 'Opponent'}), on = ['Opponent','Round'], how = 'left')
Base on your update
yourdf = df2.merge(df1.rename(columns={'ID' : 'Opponent'}), on = ['Opponent','Round'], how = 'left')
Good day SO community,
I have been having an issue with trying to highlight errors in my df, row by row.
reference_dict = {'jobclass' : ['A','B'], 'Jobs' : ['Teacher','Plumber']}
dict = {'jobclass': ['A','C','A'], 'Jobs': ['Teacher', 'Plumber','Policeman']}
df = pd.DataFrame(data=dict)
def highlight_rows(df):
for i in df.index:
if df.jobclass[i] in reference_dict['jobclass']:
print(df.jobclass[i])
return 'background-color: green'
df.style.apply(highlight_rows, axis = 1)
I am getting the error:
TypeError: ('string indices must be integers', 'occurred at index 0')
What i hope to get is my df with values not found in my reference_dict being highlighted.
Any help would be greatly appreciated.. Cheers!
Edit:
x = {'jobclass' : ['A','B'], 'Jobs' : ['Teacher','Plumber']}
d = {'jobclass': ['A','C','A'], 'Jobs': ['Teacher', 'Plumber','Policeman']}
df = pd.DataFrame(data=d)
print(df)
def highlight_rows(s):
ret = ["" for i in s.index]
for i in df.index:
if df.jobclass[i] not in x['jobclass']:
ret[s.index.get_loc('Jobs')] = "background-color: yellow"
return ret
df.style.apply(highlight_rows, axis = 1)
Tried this and got the whole column highlighted instead of the specific rows values that i desire.. =/
You can use merge with parameter indicator for found not matched values and then create DataFrame of styles:
x = {'jobclass' : ['A','B'], 'Jobs' : ['Teacher','Plumber']}
d = {'jobclass': ['A','C','A'], 'Jobs': ['Teacher', 'Plumber','Policeman']}
df = pd.DataFrame(data=d)
print (df)
jobclass Jobs
0 A Teacher
1 C Plumber
2 A Policeman
Detail:
print (df.merge(pd.DataFrame(x) , on='jobclass', how='left', indicator=True))
jobclass Jobs_x Jobs_y _merge
0 A Teacher Teacher both
1 C Plumber NaN left_only
2 A Policeman Teacher both
def highlight_rows(s):
c1 = 'background-color: yellow'
c2 = ''
df1 = pd.DataFrame(x)
m = s.merge(df1, on='jobclass', how='left', indicator=True)['_merge'] == 'left_only'
df2 = pd.DataFrame(c2, index=s.index, columns=s.columns)
df2.loc[m, 'Jobs'] = c1
return df2
df.style.apply(highlight_rows, axis = None)
Good day to you as well!
What i hope to get is my df with values not found in my reference_dict being highlighted.
If you're looking for values not found in reference_dict to be highlighted, do you mean for the function to be the following?
def highlight_rows(df):
for i in df.index:
if df.jobclass[i] not in reference_dict['jobclass']:
print(df.jobclass[i])
return 'background-color: green'
Either way, why highlight the rows when you could isolate them? It seems like you want to look at all of the job classes in df where there is not one in reference_dict.
import pandas as pd
reference_dict = {'jobclass' : ['A','B'], 'Jobs' : ['Teacher','Plumber']}
data_dict = {'jobclass': ['A','C','A'], 'Jobs': ['Teacher', 'Plumber','Policeman']}
ref_df = pd.DataFrame(reference_dict)
df = pd.DataFrame(data_dict)
outliers = df.merge(ref_df, how='outer', on='jobclass') # merge the two tables together, how='outer' includes jobclasses which the DataFrames do not have in common. Will automatically generate columns Jobs_x and Jobs_y once joined together because the columns have the same name
outliers = outliers[ outliers['Jobs_y'].isnull() ] # Jobs_y is null when there is no matching jobclass in the reference DataFrame, so we can take advantage of that by filtering
outliers = outliers.drop('Jobs_y', axis=1) # let's drop the junk column after we used it to filter for what we wanted
print("The reference DataFrame is:")
print(ref_df,'\n')
print("The input DataFrame is:")
print(df,'\n')
print("The result is a list of all the jobclasses not in the reference DataFrame and what job is with it:")
print(outliers)
The result is:
The reference DataFrame is:
jobclass Jobs
0 A Teacher
1 B Plumber
The input DataFrame is:
jobclass Jobs
0 A Teacher
1 C Plumber
2 A Policeman
The result is a list of all the jobclasses not in the reference DataFrame and what job is with it:
jobclass Jobs_x
2 C Plumber
This could have been a tangent but it's what I'd do. I was not aware you could highlight rows in pandas at all, cool trick.
I have 3 dataframes, containing daily data: unique code, names, scores. First value in Row 1 is called Rank and then I have dates, first column under Rank contains the rank number (the first column is used as index).
**df1** UNIQUE CODES
Rank 12/8/2017 12/9/2017 .... 1/3/2018
1 Code_1 Code_3 Code_4
2 Code_2 Code_1 Code_2
...
1000 Code_5 Code_6 Code_7
**df2** NAMES
Rank 12/8/2017 12/9/2017 .... 1/3/2018
1 Jon Maria Peter
2 Brian Jon Maria
...
1000 Chris Tim Charles
**df3** SCORES
Rank 12/8/2017 12/9/2017 .... 1/3/2018
1 10 20 30
2 15 10 40
...
1000 25 15 20
Desired output:
I want to combine these dataframes into a dictionary, using df1 codenames as keys, so it will look like this:
dictionary = {'Code_1':[Jon, 20] , 'Code_2':[Brian, 15]}
As there are repeat competitors, I will need to sum their scores during all of the data series. So in the above examples, the Score_1 of Jon will contain scores for 12/8/2017 and 12/9/2017.
There are 1000 rows and 26 columns + index, so need a way to capture those. I think that a nested loop could work here, but don't have enough experience to build one that works.
In the end, I would like to sort the dictionary by highest score. Please suggest any solutions to this or more straightforward ways to combine this data and get the score ranking.
I attached pictures of dataframes, containing names, codes, and scores.
names
codes
scores
I used the proposed solution below on the 3 dataframes that I have. Please note that hashtags stands for code, players for names, and trophies for scores:
# reshape to get dates into rows
hashtags_reshaped = pd.melt(hashtags, id_vars = ['Rank'],
value_vars = hashtags.columns,
var_name = 'Date',
value_name = 'Code').drop('Rank', axis = 1)
# reshape to get dates into rows
players_reshaped = pd.melt(players, id_vars = ['Rank'],
value_vars = hashtags.columns,
var_name = 'Date',
value_name = 'Name').drop('Rank', axis = 1)
# reshape to get the dates into rows
trophies_reshaped = pd.melt(trophies, id_vars = ['Rank'],
value_vars = hashtags.columns,
var_name = 'Date',
value_name = 'Score').drop('Rank', axis = 1)
# merge the three together.
# This _assumes_ that the dfs are all in the same order and that all the data matches up.
merged_df = pd.DataFrame([hashtags_reshaped['Date'],
hashtags_reshaped['Code'], players_reshaped['Name'],
trophies_reshaped['Score']]).T
print(merged_df)
# group by code, name, and date; sum the scores together if multiple exist for a given code-name-date grouping
grouped_df = merged_df.groupby(['Code', 'Name', 'Date']).sum().sort_values('Score', ascending = False)
print(grouped_df)
summed_df = merged_df.drop('Date', axis = 1) \
.groupby(['Code', 'Name']).sum() \
.sort_values('Score', ascending = False).reset_index()
summed_df['li'] = list(zip(summed_df.Name, summed_df.Score))
print(summed_df)
But I'm getting a strange output: the summed scores should be in hundreds or low thousands (as an average score is 200-300 and an average participation frequency is 4-6 times). The score results I'm getting are way off, but their match codes and names correctly.
summed_df:
0 (MandiBralaX, 996871590076253)
1 (Arso_C, 9955130513430)
2 (ThatRainbowGuy, 9946)
3 (fabi, 9940)
4 (Dogão, 991917)
5 (Hierbo, 99168)
6 (Clyde, 9916156180128)
7 (.A.R.M.I.N., 9916014310187143)
8 (keftedokofths, 9900)
9 (⚽AngelSosa⚽, 990)
10 (Totoo98, 99)
group_df:
Code Name Score \
0 #JL2J02LY MandiBralaX 996871590076253
1 #80JQ90VC Arso_C 9955130513430
2 #9GGC2CUQ ThatRainbowGuy 9946
3 #8LL989QV fabi 9940
4 #9PPC89L Dogão 991917
5 #2JPLQ8JP8 Hierbo 99168
This should get you much of the way there. I didn't create a dictionary at the end as you specified; while you may need that format, you'd end up with nested dictionaries or lists, as each Code has 1 Name but possibly many Dates and Scores associated with it. How do you want those recorded - list, dict, etc?
The code below returns a grouped dataframe; you can output it directly to a dict (shown), but you'll probably want to specify the format in detail, especially if you need an ordered dictionary. (Dictionaries are inherently not ordered; you'll have to from collections import OrderedDict and review that documentation if you really need an ordered dictionary.
import pandas as pd
#create the dfs; note that 'Code' is set up as a string
df1 = pd.DataFrame({'Rank': [1, 2], '12/8/2017': ['1', '2'], '12/9/2017': ['3', '1']})
df1.set_index('Rank', inplace = True)
# reshape to get dates into rows
df1_reshaped = pd.melt(df1, id_vars = ['Rank'],
value_vars = df1.columns,
var_name = 'Date',
value_name = 'Code').drop('Rank', axis = 1)
#print(df1_reshaped)
# create the second df
df2 = pd.DataFrame({'Rank': [1, 2], '12/8/2017': ['Name_1', 'Name_2'], '12/9/2017': ['Name_3', 'Name_1']})
df2.set_index('Rank', inplace = True)
# reshape to get dates into rows
df2_reshaped = pd.melt(df2, id_vars = ['Rank'],
value_vars = df1.columns,
var_name = 'Date',
value_name = 'Name').drop('Rank', axis = 1)
#print(df2_reshaped)
# create the third df
df3 = pd.DataFrame({'Rank': [1, 2], '12/8/2017': ['10', '20'], '12/9/2017': ['30', '10']})
df3.set_index('Rank', inplace = True)
# reshape to get the dates into rows
df3_reshaped = pd.melt(df3, id_vars = ['Rank'],
value_vars = df1.columns,
var_name = 'Date',
value_name = 'Score').drop('Rank', axis = 1)
#print(df3_reshaped)
# merge the three together.
# This _assumes_ that the dfs are all in the same order and that all the data matches up.
merged_df = pd.DataFrame([df1_reshaped['Date'], df1_reshaped['Code'], df2_reshaped['Name'], df3_reshaped['Score']]).T
print(merged_df)
# group by code, name, and date; sum the scores together if multiple exist for a given code-name-date grouping
grouped_df = merged_df.groupby(['Code', 'Name', 'Date']).sum().sort_values('Score', ascending = False)
print(grouped_df)
summed_df = merged_df.drop('Date', axis = 1) \
.groupby(['Code', 'Name']).sum() \
.sort_values('Score', ascending = False).reset_index()
summed_df['li'] = list(zip(summed_df.Name, summed_df.Score))
print(summed_df)
Unsorted dict:
d = dict(zip(summed_df.Code, summed_df.li))
print(d)
You can make the OrderedDict directly, of course, and should:
from collections import OrderedDict
d2 = OrderedDict(zip(summed_df.Code, summed_df.li))
print(d2)
summed_df:
Code Name Score li
0 3 Name_3 30 (Name_3, 30)
1 1 Name_1 20 (Name_1, 20)
2 2 Name_2 20 (Name_2, 20)
d:
{'3': ('Name_3', 30), '1': ('Name_1', 20), '2': ('Name_2', 20)}
d2, sorted:
OrderedDict([('3', ('Name_3', 30)), ('1', ('Name_1', 20)), ('2', ('Name_2', 20))])
This returns your (name, score) as a tuple, not a list, but... it should get more of the way there.