We Keep Coding

Cumcount with reset in Python

I have a following problem. I need to compute a cumcount but I would like to reset the counter always when the series is interupted.
See example:
data = { 'col_1': ['a', 'a', 'b', 'b', 'a'], 'col_2': [3, 2, 1, 0, -3]}
df = pd.DataFrame.from_dict(data)
I tried this but it gives me a wrong output:
df["seq"] = df.groupby(["col_1"]).cumcount()
What I want is:
data = { 'col_1': ['a', 'a', 'b', 'b', 'a'], 'col_2': [3, 2, 1, 0, -3], 'seq': [0, 1, 0, 1, 0]}
How can I do it, please?

Try:
df["seq"] = df.groupby((df["col_1"] != df["col_1"].shift()).cumsum())["col_1"].cumcount()
print(df)
Output
col_1 col_2 seq
0 a 3 0
1 a 2 1
2 b 1 0
3 b 0 1
4 a -3 0

Note that as you are interested in runs (like in run-length encoding) itertools.groupby might be better suited for this task, consider following example
import pandas as pd
df = pd.DataFrame({'col1':['a','a','b','b','a']})
df['seq'] = [i for k, g in itertools.groupby(col1) for i in range(len(list(g)))]
print(df)
output
col1 seq
0 a 0
1 a 1
2 b 0
3 b 1
4 a 0

Replace values of a dataframe with the value of another dataframe

I have two pandas dataframes
df1 = pd.DataFrame({'A': [1, 3, 5], 'B': [3, 4, 5]})
df2 = pd.DataFrame({'A': [1, 2, 3, 4, 5], 'B': [8, 9, 10, 11, 12], 'C': ['K', 'D', 'E', 'F', 'G']})
The index of both data-frames are 'A'.
How to replace the values of df1's column 'B' with the values of df2 column 'B'?
RESULT of df1:
A B
1 8
3 10
5 12

Maybe dataframe.isin() is what you're searching:
df1['B'] = df2[df2['A'].isin(df1['A'])]['B'].values
print(df1)
Prints:
A B
0 1 8
1 3 10
2 5 12

One of possible solutions:
wrk = df1.set_index('A').B
wrk.update(df2.set_index('A').B)
df1 = wrk.reset_index()
The result is:
A B
0 1 8
1 3 10
2 5 12
Another solution, based on merge:
df1 = df1.merge(df2[['A', 'B']], how='left', on='A', suffixes=['_x', ''])\
.drop(columns=['B_x'])

groupby in pandas with function that must keep state

I have the following dataframe
df = pd.DataFrame({'a': ['A', 'A', 'A', 'B', 'B', 'B', 'B'],
'b': [ 1, 2, 4, 1, 2, 3, 4]})
I want a function that would output the following dataframe definition:
df = pd.DataFrame({'a': [ 'A', 'A', 'A', 'B', 'B', 'B', 'B'],
'b': [ 1, 2, 4, 1, 2, 3, 4],
'c': ['A_0', 'A_0', 'A_1', 'B_0', 'B_0', 'B_0', 'B_0']})
The logic is that given for each value of 'a' (each group), I create a value 'c' which could be described as a "continuous" series of 'b' values.
So far, my code is the following:
def detection(dataset):
def detect(series, avalue):
_id = 0
start = True
visits = []
prev_ = None
for h in series:
if start:
start = False
prev_ = h
else:
if h - prev_ > 1:
_id += 1
prev_ = h
visits.append(f"{avalue}_{_id}")
return visits
res = []
gb = dataset.groupby("a")
for avalue in gb.groups:
dd = gb.get_group(avalue)
dd["VISIT_ID"] = detect(dd["b"], avalue)
res.append(dd)
return pd.concat(res, axis=0)
The good is, it works perfectly !!
The bad: it is extremely slow on a large dataset (7 millions of entries, 250k of different 'a' values.
Is there something better to do?

You can find the numeric part of column c using groupby and concat values
df['c'] = df.groupby('a').b.apply(lambda x: (x.diff() > 1).cumsum())
df['c'] = df['a'] + '_' + df['c'].astype(str)
a b c
0 A 1 A_0
1 A 2 A_0
2 A 4 A_1
3 B 1 B_0
4 B 2 B_0
5 B 3 B_0
6 B 4 B_0

How to replace specific character in pandas column with null?

I Have a column within a dataset, regarding categorical company sizes, which currently looks like this, where the '-' hyphens are currently representing missing data:
I want to change the '-' in missing values with nulls so i can analyse missing data. However when I use the pd replace tool (see following code) with a None value it seems to also make any of the genuine entries as they also contain hyphens (e.g 51-200).
df['Company Size'].replace({'-': None},inplace =True, regex= True)
How can I replace only lone standing hyphens and leave the other entries untouched?

You need not to use regex=True.
df['Company Size'].replace({'-': None},inplace =True)

You could also just do:
df['column_name'] = df['column_name'].replace('-','None')

import numpy as np
df.replace('-', np.NaN, inplace=True)
This code worked for me.

you can do it like this
import numpy as np
import pandas as pd
df = pd.DataFrame({'A': [0, 1, 2, 3, 4],
'B': [5, 6, 7, 8, 9],
'C': ['a', '-', 'c--', 'd', 'e']})
df['C'] = df['C'].replace('-', np.nan)
df = df.where((pd.notnull(df)), None)
# can also use this -> df['C'] = df['C'].where((pd.notnull(df)), None)
print(df)
output:
A B C
0 0 5 a
1 1 6 None
2 2 7 c--
3 3 8 d
4 4 9 e
another example:
df = pd.DataFrame({'A': [0, 1, 2, 3, 4],
'B': ['5-5', '-', 7, 8, 9],
'C': ['a', 'b', 'c--', 'd', 'e']})
df['B'] = df['B'].replace('-', np.nan)
df = df.where((pd.notnull(df)), None)
print(df)
output:
A B C
0 0 5-5 a
1 1 None b
2 2 7 c--
3 3 8 d
4 4 9 e

Replicating rows in pandas dataframe by column value and add a new column with repetition index

My question is similar to one asked here. I have a dataframe and I want to repeat each row of the dataframe k number of times. Along with it, I also want to create a column with values 0 to k-1. So
import pandas as pd
df = pd.DataFrame(data={
'id': ['A', 'B', 'C'],
'n' : [ 1, 2, 3],
'v' : [ 10, 13, 8]
})
what_i_want = pd.DataFrame(data={
'id': ['A', 'B', 'B', 'C', 'C', 'C'],
'n' : [ 1, 2, 2, 3, 3, 3],
'v' : [ 10, 13, 13, 8, 8, 8],
'repeat_id': [0, 0, 1, 0, 1, 2]
})
Command below does half of the job. I am looking for pandas way of adding the repeat_id column.
df.loc[df.index.repeat(df.n)]

Use GroupBy.cumcount and copy for avoid SettingWithCopyWarning:
If you modify values in df1 later you will find that the modifications do not propagate back to the original data (df), and that Pandas does warning.
df1 = df.loc[df.index.repeat(df.n)].copy()
df1['repeat_id'] = df1.groupby(level=0).cumcount()
df1 = df1.reset_index(drop=True)
print (df1)
id n v repeat_id
0 A 1 10 0
1 B 2 13 0
2 B 2 13 1
3 C 3 8 0
4 C 3 8 1
5 C 3 8 2