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Python Pandas: Subsetting gives back Series instead of integer [duplicate]

I have dataframe df_my that looks like this
id name age major
----------------------------------------
0 1 Mark 34 English
1 2 Tom 55 Art
2 3 Peter 31 Science
3 4 Mohammad 23 Math
4 5 Mike 47 Art
...
I am trying to get the value of major (only)
I used this and it works fine when I know the id of the record
df_my["major"][3]
returns
"Math"
great
but I want to get the major for a variable record
I used
i = 3
df_my.loc[df_my["id"]==i]["major"]
and also used
i = 3
df_my[df_my["id"]==i]["major"]
but they both return
3 Math
it includes the record index too
how can I get the major only and nothing else?

You could use squeeze:
i = 3
out = df.loc[df['id']==i,'major'].squeeze()
Another option is iat:
out = df.loc[df['id']==i,'major'].iat[0]
Output:
'Science'

I also stumbled over this problem, from a little different angle:
df = pd.DataFrame({'First Name': ['Kumar'],
'Last Name': ['Ram'],
'Country': ['India'],
'num_var': 1})
>>> df.loc[(df['First Name'] == 'Kumar'), "num_var"]
0 1
Name: num_var, dtype: int64
>>> type(df.loc[(df['First Name'] == 'Kumar'), "num_var"])
<class 'pandas.core.series.Series'>
So it returns a Series (although it is only a series with only 1 element). If you access through the index, you receive the integer.
df.loc[0, "num_var"]
1
type(df.loc[0, "num_var"])
<class 'numpy.int64'>
The answer on how to select the respective, single value was already given above. However, I think it is interesting to note that accessing through an index always gives the single value whereas accessing through a condition returns a series. This is, b/c accessing with index clearly returns only one value whereas accessing through a condition can return several values.

If one of the columns of your dataframe is the natural primary index for those data, then it's usually a good idea to make pandas aware of it by setting the index accordingly:
df_my.set_index('id', inplace=True)
Now you can easily get just the major value for any id value i:
df_my.loc[i, 'major']
Note that for i = 3, the output is 'Science', which is expected, as noted in the comments to your question above.

Finding number of survived people in Titanic Dataset in Python

From the Titanic Dataset from Kaggle, I'm trying to extract how many people survived and how many died from the survived column. To do this, I imported the pandas library and saved the dataset in the variable dataframe and used the following code:
dataframe['survived'].value_counts()
Which gave me the output as
0 809
1 500
Name: survived, dtype: int64
From this, how do I print just the number of people who survived? Like if I want the count of 1, I need the output as 500. Same thing for when I want just the count of 0.
I tried the following code only to get a SyntaxError
dataframe['survived'].value_counts().1
I'm new to pandas, so I'd really appreciate it if anyone could help me with this!

For your case, you can use sum instead of value_counts because you have a binary column: 1 for survived, 0 for died so the sum get you all survived people:
>>> dataframe['survived'].sum()
500
In case of your column is not binary, you can use:
# 1 stand for survived people here
>>> dataframe['survived'].eq(1).sum()
500

Here is a more human like logic answer.
Ask the data frame for all observations (rows) with survivors in it.
# some datasets would use 'yes', 'si', 'alive', ..
alive = 1
# eq() means equal; like ==
survivors = dataframe[dataframe.survived.eq(alive)]
And then count the observations (rows).
print(len(survivors))

You can use:
dataframe['survived'].value_counts()[0]
or:
dataframe['survived'].value_counts().loc[0]
The .column_name/.index_name syntax is not recommended as it restricts the possibilities to column names that are valid python variables. Strings starting with a number are not valid python variable names.

On my database, I got an attribute that has a lot of diferent values separeted with comma. E.G: Attribute "sintomas" - value "Outros,tosse"

I got a project on my college that we have to take an database an make some data analasys on it.
My problem is that the database has one attribute that have a lot of different datas on in, separeted with comma.
I have to do the KNN algorithym on it and I'm changing all the qualitative values to numbers, but I dont know what to do on the situation showed below. I was thinking on getting all the substrings and adding as attribute, for example, in the row 0 i got "Outros, Tosse", so i would have an attribute "Outros" and another "Tosse", and if the substring is present, the value would be "1" and if its not, the value would be "0". I'm using colab with pandas right now. Someone knows what can i do in this situation?
Data sample:
sintomas profissionalsaude
0 Outros, Tosse 2
1 Febre, Tosse, Distúrbios Gustativos 1
.
.
.
Thanks!

You can do something like this:
pd.get_dummies(
df['sintomas'].str.split(', ').explode()
).groupby(level=0).sum()
The idea is to first split strings by ,, then explode it into a column with repeating indexes (with index representing the original index = row number), then get_dummies to do one-hot encoding, groupby that index and sum
For example:
df = pd.DataFrame({'feature': ['a,b, c', 'b, c', 'a']})
pd.get_dummies(
df['feature'].str.split(',\s*').explode()
).groupby(level=0).sum()
Output:
a b c
0 1 1 1
1 0 1 1
2 1 0 0

Python, lambda function as argument for groupby

I'm trying to figure out what a piece of code is doing, but I'm getting kinda lost on it.
I have a pandas dataframe, which has been loaded by the following .csv file:
origin_census_block_group,date_range_start,date_range_end,device_count,distance_traveled_from_home,bucketed_distance_traveled,median_dwell_at_bucketed_distance_traveled,completely_home_device_count,median_home_dwell_time,bucketed_home_dwell_time,at_home_by_each_hour,part_time_work_behavior_devices,full_time_work_behavior_devices,destination_cbgs,delivery_behavior_devices,median_non_home_dwell_time,candidate_device_count,bucketed_away_from_home_time,median_percentage_time_home,bucketed_percentage_time_home,mean_home_dwell_time,mean_non_home_dwell_time,mean_distance_traveled_from_home
010539707003,2020-06-25T00:00:00-05:00,2020-06-26T00:00:00-05:00,49,626,"{""16001-50000"":5,""0"":11,"">50000"":4,""2001-8000"":3,""1-1000"":9,""1001-2000"":7,""8001-16000"":1}","{""16001-50000"":110,"">50000"":155,""<1000"":40,""2001-8000"":237,""1001-2000"":27,""8001-16000"":180}",12,627,"{""721-1080"":11,""361-720"":9,""61-360"":1,""<60"":11,"">1080"":12}","[32,32,28,30,30,31,27,23,20,20,20,17,19,19,15,14,17,20,20,21,25,22,24,23]",7,3,"{""120330012011"":1,""010030107031"":1,""010030114052"":2,""120330038001"":1,""010539701003"":1,""010030108001"":1,""010539707002"":14,""010539705003"":2,""120330015001"":1,""121130102003"":1,""010539701002"":1,""120330040001"":1,""370350101014"":2,""120330033081"":2,""010030106003"":1,""010539706001"":2,""010539707004"":3,""120330039001"":1,""010539699003"":1,""120330030003"":1,""010539707003"":41,""010970029003"":1,""010539705004"":1,""120330009002"":1,""010539705001"":3,""010539704003"":1,""120330028012"":1,""120330035081"":1,""120330036102"":1,""120330036142"":1,""010030114062"":1,""010539706004"":7,""010539706002"":1,""120330036082"":1,""010539707001"":7,""010030102001"":1,""120330028011"":1}",2,241,71,"{""21-45"":4,""481-540"":2,""541-600"":1,""721-840"":1,""1201-1320"":1,""301-360"":3,""<20"":13,""61-120"":3,""241-300"":3,""121-180"":1,""421-480"":3,""1321-1440"":4,""1081-1200"":1,""961-1080"":2,""601-660"":1,""181-240"":1,""661-720"":2,""361-420"":3}",72,"{""0-25"":13,""76-100"":21,""51-75"":6,""26-50"":3}",657,413,1936
010730144081,2020-06-25T00:00:00-05:00,2020-06-26T00:00:00-05:00,139,2211,"{""16001-50000"":17,""0"":41,"">50000"":15,""2001-8000"":22,""1-1000"":8,""1001-2000"":12,""8001-16000"":24}","{""16001-50000"":143,"">50000"":104,""<1000"":132,""2001-8000"":39,""1001-2000"":15,""8001-16000"":102}",41,806,"{""721-1080"":32,""361-720"":16,""61-360"":12,""<60"":30,"">1080"":46}","[91,92,93,91,91,90,86,83,78,64,64,61,64,62,65,62,60,74,61,64,75,78,81,84]",8,6,"{""131350501064"":1,""131350502151"":1,""010730102002"":1,""011170302131"":2,""010730038024"":1,""010730108041"":1,""010730144133"":1,""010730132003"":1,""011210118002"":1,""011170303053"":1,""010730111084"":2,""011170302142"":1,""010730119011"":1,""010730129063"":2,""010730107063"":1,""010730059083"":1,""010730058003"":1,""011270204003"":1,""010730049012"":2,""130879701001"":1,""010730120021"":1,""130890219133"":1,""010730144082"":4,""170310301031"":1,""010730129112"":1,""010730024002"":1,""011170303034"":2,""481390616004"":1,""121270826052"":1,""010730128021"":2,""121270825073"":1,""010730004004"":1,""211959313002"":1,""010730100012"":1,""011170302151"":1,""010730142041"":1,""010730129123"":1,""010730129084"":1,""010730042002"":1,""010730059033"":2,""170318306001"":1,""130519800001"":1,""010730027003"":1,""121270826042"":1,""481610001002"":1,""010730100011"":1,""010730023032"":1,""350250004002"":1,""010730056003"":1,""010730132001"":1,""011170302171"":2,""120910227003"":1,""011239620001"":1,""130351503002"":1,""010730129155"":1,""010730001001"":2,""010730110021"":1,""170310104003"":1,""010730059082"":2,""010730120022"":1,""011170303151"":1,""010730139022"":1,""011170303441"":4,""010730144092"":3,""010730129151"":1,""011210119001"":2,""010730144081"":117,""010730108052"":1,""010730129122"":9,""370710321003"":1,""010730142034"":2,""010730042001"":2,""010570201003"":1,""010730144132"":6,""010730059032"":1,""010730012001"":2,""010730102003"":1,""011170303332"":1,""010730128032"":2,""010730129081"":1,""010730103011"":1,""010730058001"":3,""011150401041"":1,""010730045001"":3,""010730110013"":1,""010730119041"":1,""010730042003"":1,""010730141041"":1,""010730144091"":1,""010730129154"":1,""484759501002"":1,""010730144063"":1,""010730144102"":12,""011170303141"":1,""011250106011"":1,""011170303152"":1,""010730059104"":1,""010730107021"":1,""010730100014"":1,""010730008004"":1,""011170303451"":1,""010730127041"":2,""370559704003"":1,""010730047011"":2,""010730129132"":2,""011010014002"":1,""010730144131"":1,""011170302133"":1,""010730030011"":1,""131350506063"":1,""010730118023"":1,""010890110141"":1,""010730128023"":1,""010730106022"":2,""130879703004"":1,""010730108015"":1,""131390010041"":1,""011170305013"":1,""010730134002"":1,""010730031004"":1,""010730138012"":1,""010730011004"":1,""011250102041"":1,""010730129131"":4,""010730144101"":4,""011170303331"":2,""010730003001"":1,""011010033012"":1,""483539504004"":1,""010550104021"":1,""011170303411"":1,""010730106031"":1,""011170303153"":5,""010730128034"":1,""010730129061"":1,""131390010023"":1,""010730051042"":1,""130510107002"":1,""010730027001"":2,""120090686011"":1,""010730107042"":1,""010730123052"":1,""010730129102"":1,""011210115003"":1,""010730129083"":4,""011170303142"":1,""011010014001"":1,""010730107064"":2}",7,176,205,"{""21-45"":7,""481-540"":10,""541-600"":4,""46-60"":2,""721-840"":3,""1201-1320"":3,""301-360"":7,""<20"":46,""61-120"":6,""241-300"":4,""121-180"":9,""421-480"":2,""1321-1440"":3,""1081-1200"":5,""961-1080"":1,""601-660"":1,""181-240"":5,""661-720"":1,""361-420"":7}",78,"{""0-25"":29,""76-100"":71,""51-75"":27,""26-50"":8}",751,338,38937
010890017002,2020-06-25T00:00:00-05:00,2020-06-26T00:00:00-05:00,78,1934,"{""16001-50000"":2,""0"":12,"">50000"":9,""2001-8000"":27,""1-1000"":12,""1001-2000"":8,""8001-16000"":8}","{""16001-50000"":49,"">50000"":99,""<1000"":111,""2001-8000"":37,""1001-2000"":24,""8001-16000"":28}",11,787,"{""721-1080"":17,""361-720"":11,""61-360"":11,""<60"":15,"">1080"":23}","[49,42,48,48,47,48,44,44,39,32,34,32,36,31,32,36,40,37,36,38,49,45,46,46]",5,1,"{""010890101002"":1,""010730108041"":1,""010890020003"":2,""010890010001"":2,""010890025011"":3,""010890026001"":4,""280819505003"":1,""281059504004"":1,""010890103022"":1,""120990056011"":1,""010890109012"":2,""010890019021"":6,""010890013021"":4,""010890015004"":3,""010890108003"":1,""010890014022"":6,""281059501003"":1,""281059503001"":1,""010890007022"":3,""010890017001"":3,""010890107023"":1,""010890021002"":1,""010890009011"":1,""010890109013"":1,""010730120022"":1,""010890031003"":15,""011170303151"":1,""010890019011"":9,""010890030002"":2,""010890110221"":1,""011170305021"":1,""010890026003"":2,""010890025012"":3,""010730117034"":1,""010830208022"":1,""010890031002"":2,""010890112002"":1,""010210602001"":1,""010890002022"":1,""010890017002"":65,""281059506021"":1,""010890010003"":2,""010890106222"":1,""120990059182"":1,""010890110222"":1,""010890020001"":1,""010890101003"":1,""010890018013"":1,""010890021001"":1,""010890109021"":1,""010890108001"":1,""010770106005"":1,""281059506011"":1,""010030114032"":2,""010830209001"":1,""010890027222"":1,""010730128023"":1,""010890009021"":1,""010030114051"":1,""010030109031"":1,""010030103003"":1,""010890031001"":1,""010890021003"":1,""010030114062"":4,""010890106241"":1,""281059504003"":1,""010890018011"":10,""010890019031"":5,""010890027012"":1,""010730108054"":1,""010890106223"":2,""010890111001"":1,""010210603002"":1,""010890109011"":1,""010890019012"":2,""010890113001"":1,""010890028013"":3}",1,229,99,"{""481-540"":3,""541-600"":2,""46-60"":1,""721-840"":1,""1201-1320"":7,""301-360"":6,""<20"":18,""61-120"":10,""241-300"":5,""121-180"":2,""1321-1440"":2,""841-960"":1,""1081-1200"":1,""961-1080"":3,""601-660"":3,""181-240"":2,""661-720"":3}",78,"{""0-25"":16,""76-100"":44,""51-75"":11,""26-50"":7}",708,353,14328
010950308022,2020-06-25T00:00:00-05:00,2020-06-26T00:00:00-05:00,100,2481,"{""16001-50000"":11,""0"":19,"">50000"":11,""2001-8000"":40,""1-1000"":6,""1001-2000"":3,""8001-16000"":4}","{""16001-50000"":150,"">50000"":23,""<1000"":739,""2001-8000"":23,""1001-2000"":12,""8001-16000"":208}",17,703,"{""721-1080"":21,""361-720"":19,""61-360"":10,""<60"":24,"">1080"":26}","[62,64,64,63,65,67,54,48,37,37,34,33,30,34,32,33,35,43,50,56,58,56,56,57]",8,6,"{""010950306004"":1,""010950302023"":1,""011030054051"":1,""010950311002"":1,""010950309023"":1,""010499606003"":1,""121319506023"":2,""010950308022"":86,""121319506016"":2,""010950304013"":1,""010950307024"":1,""010950309041"":1,""010890019021"":2,""010950312001"":5,""010499607002"":1,""011150402013"":1,""010550102003"":1,""120050027043"":3,""010719509003"":1,""010950302022"":1,""010950308023"":2,""120050027051"":2,""471079701022"":1,""010890106221"":1,""010950306001"":1,""010950302011"":2,""011150405013"":1,""011150402041"":2,""010950312002"":16,""011030054042"":1,""010950301002"":2,""130459105011"":1,""010730001001"":1,""130459102001"":1,""010890109013"":2,""010950308013"":14,""010719508004"":1,""120050027041"":3,""010550110021"":3,""010730049022"":1,""010950308024"":1,""010950312004"":6,""010950312003"":1,""010550104012"":2,""010550110013"":1,""120860004111"":1,""010890027222"":1,""010950306002"":2,""010950304015"":1,""011030054041"":1,""010950309031"":8,""010950308021"":1,""010950302024"":1,""010950307011"":5,""010550110012"":2,""011150404013"":1,""130459103003"":1,""120050027032"":3,""010950307012"":5,""010950309022"":2,""010950307023"":1,""010719508003"":1,""010499608001"":2,""010950310003"":1,""011150402043"":1,""120860099063"":1,""010950309021"":4,""010950309043"":2,""010950308011"":1,""010950306003"":3,""120050027042"":1,""010950308025"":5,""010950309032"":6,""010499607001"":1}",1,199,132,"{""21-45"":8,""481-540"":6,""541-600"":4,""46-60"":3,""721-840"":3,""1201-1320"":4,""301-360"":3,""<20"":20,""61-120"":10,""241-300"":2,""121-180"":4,""421-480"":3,""1321-1440"":1,""841-960"":3,""961-1080"":2,""601-660"":1,""181-240"":3,""661-720"":1,""361-420"":2}",74,"{""0-25"":20,""76-100"":48,""51-75"":23,""26-50"":4}",661,350,5044
df = pd.read_csv(csv_file,
usecols=[
'origin_census_block_group',
'date_range_start',
'date_range_end',
'device_count',
'distance_traveled_from_home',
'completely_home_device_count',
'median_home_dwell_time',
'part_time_work_behavior_devices',
'full_time_work_behavior_devices'
],
dtype={'origin_census_block_group': str},
).set_index('origin_census_block_group')
and, later in the code, the dataframe is modified by:
df = df.groupby(lambda cbg: cbg[:5]).sum()
I don't quite understand what this line is doing precisely.
Groupby generally groups a dataframe by column, so...is it grouping the dataframe using multiple columns (0 to 5)? What is the effect of .sum() at the end?

If you run your code exactly as you wrote it (both the creation of df and the groupby) you can see the result. I print first couple of columns of the output of groupby
device_count distance_traveled_from_home
----- -------------- -----------------------------
01053 49 626
01073 139 2211
01089 78 1934
01095 100 2481
What happens here is the function lambda cbg: cbg[:5] is applied to each of the index values (strings that look like numbers in column origin_census_block_group). As a side, note the statement
...
dtype={'origin_census_block_group': str},
when creating the df, so somebody went into trouble to make sure they are actually str
So the function is applied to string like '010539707003' and returns a substring which is the first 5 characters of that string:
'010539707003'[:5]
produces
'01053'
so I assume there are multiple keys that share the first 5 characters (in the actual file -- the snippet has them all unique so not very interesting) and all these rows are grouped together
Then .sum() is applied to each numerical column of each group and returns, well, the column sum per each groupby key. This is what you see in my output in column 'device_count' and so on.
Hope this is clear now

Pandas' read_csv() will render a csv-formatted file a Pandas Dataframe.
I recommend having a ready at the Pandas' documentation, as it's very exhaustive -> https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.read_csv.html
usecols=[
'origin_census_block_group',
'date_range_start',
'date_range_end',
'device_count',
'distance_traveled_from_home',
'completely_home_device_count',
'median_home_dwell_time',
'part_time_work_behavior_devices',
'full_time_work_behavior_devices'
],
The usecols parameter will take as input an array of desired columns and will only load the specified columns into the dataframe.
dtype={'origin_census_block_group': str}
The dtype parameter will take a dict as input and is to specify the data type of the values, like {'column' : datatype}
.set_index('origin_census_block_group')
.set_index() will set the specificed column as the index column (ie: the first column). The usual index of Pandas' Dataframe is the row's index number, which appears as the first column of the dataframe. By setting the index, the first column now becomes the specified column. See: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.set_index.html
Panda's .groupby() function will take a dataframe a regroup it basing on the occurrences of he values from the specified column.
That is to say, if we a dataframe such as df =
Fruit Name Quality Count
Apple Marco High 4
Pear Lucia Medium 10
Apple Francesco Low 3
Banana Carlo Medium 6
Pear Timmy Low 7
Apple Roberto High 8
Banana Joe High 21
Banana Jack Low 3
Pear Rob Medium 5
Apple Louis Medium 6
Pear Jennifer Low 7
Pear Laura High 8
Performing a groupby operations, such as:
df = df.groupby(lambda x: x[:2]).sum()
Will take all the elements in the index, slice them from index 0 through index 2 and return the sum of all the corresponding values, ie:
Ap 21
Ba 30
Pe 37
Now, you might be wondering about that final .sum() method. If you try to print the dataframe without applying it, you'll likely get something like this:
<bound method GroupBy.sum of <pandas.core.groupby.generic.DataFrameGroupBy object at 0x109d260a0>>
This is because Pandas has created a groubpy object and does not yet now how to display it to you. Do you want to have it displayed by the number of the occurrences in the index? You'd do this:
df = df.groupby(lambda x: x[:2]).size()
And that would output:
Ap 4
Ba 3
Pe 5
Or maybe the sum of their respective summable values? (Which is what is done in the example)
df = df.groupby(lambda x: x[:2]).sum()
Which again, will output:
Ap 21
Ba 30
Pe 37
Notice it has taken the first two letters of the string in the index. Had it been x[:3], it would have taken the first three letters, of course.
Summing it up:
-> .groupby() takes the elements in the index, i.e. the first column of the dataframe and organises the dataframe in groups relating to the index
-> The input you have given to groubpy is an anonymous function, i.e. lambda function, slicing from index 0 through 5 of its mapped input
-> You may choose how to have the results of groubpy by appending the methos .sum() or .size() to a groubpy object
I also recommend reading about Python's lambda functions:
https://docs.python.org/3/reference/expressions.html

How to split a column with string values like '[title:item]['title2:item]'...etc into a dictionary with pandas

I am trying to clean some data in a dataframe. In particular a column that displays like this:
0 [Bean status:Whole][Type of Roast:Medium][Coff...
1 [Type of Roast:Espresso][Coffee Type:Blend]
2 [Bean status:Whole][Type of Roast:Dark][Coffee...
3 [Bean status:Whole][Type of Roast:Light][Coffe...
4 NaN
5 [Roaster:Little City][Type of Roast:Light][Cof...
Name: options, dtype: object
My goal is to split this into four columns and assign the corresponding value to the columns to look something like this:
Roaster Bean Status Type of Roast Coffee Type
0 NaN Whole Medium Blend
1 NaN NaN Espresso Blend
..
5 Littl... Whole Light Single Origin
I've tried df.str.split('[', expand=True) but it is not suitable because the options are not always present or in the same position.
My thoughts were to try to split the strings into a dictionary and store that dictionary in a new dataframe, then join the two dataframes together. However, I'm getting lost trying to store the column into a dictionary. I tried doing this: https://www.fir3net.com/Programming/Python/python-split-a-string-into-a-dictionary.html like so:
roasts = {}
roasts = dict(x.split(':') for x in df['options'][0].split('[]'))
print(roasts)
and I get this error:
ValueError: dictionary update sequence element #0 has length 4; 2 is required
I tried investigating what was going on here by storing to a list instead:
s = ([x.split(':') for x in df['options'][0].split('[]')])
print(s)
[['[Bean status', 'Whole][Type of Roast', 'Medium][Coffee Type', 'Blend]']]
So I see the code is not splitting the string up how I would like, and have played around substituting a single bracket into those various locations without proper results.
Is it possible to get this column into a dictionary or will I have to resort to regex?

Using AmiTavory's sample data
df = pd.DataFrame(dict(options=[
'[Bean status:Whole][Type of Roast:Medium]',
'[Type of Roast:Espresso][Coffee Type:Blend]'
]))
Combination of re.findall and str.split
import re
import pandas as pd
pd.DataFrame([
dict(
x.split(':')
for x in re.findall('\[(.*?)\]', v)
)
for v in df.options
])
Bean status Coffee Type Type of Roast
0 Whole NaN Medium
1 NaN Blend Espresso

You might use
df.options.apply(
lambda s: pd.Series({e.split(':')[0]: e.split(':')[1] for e in s[1: -1].split('][')}))
Example
df = pd.DataFrame(dict(options=[
'[Bean status:Whole][Type of Roast:Medium]',
'[Type of Roast:Espresso][Coffee Type:Blend]'
]))
>>> df.options.apply(
lambda s: pd.Series({e.split(':')[0]: e.split(':')[1] for e in s[1: -1].split('][')}))
Bean status Coffee Type Type of Roast
0 Whole NaN Medium
1 NaN Blend Espresso
Explanation
Say you start with a string like
s = '[Bean status:Whole][Type of Roast:Medium]'
Then
s[1: -1]
removes the first and last parentheses.
Then,
split('][')
splits the dividers
Then,
e.split(':')[0]: e.split(':')[1]
for each of the splits, maps the first part to the second part.
Finally, create a Series from this.