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Python iloc slice range from dictionary value

I am trying to use a dictionary value to define the slice ranges for the iloc function but I keep getting the error -- Can only index by location with a [integer, integer slice (START point is INCLUDED, END point is EXCLUDED), listlike of integers, boolean array] . The excel sheet is built for visual information and not in any kind of real table format (not mine so I can’t change it) so I have to slice the specific ranges without column labels.
tried code - got the error
cr_dict= {'AA':'[42:43,32:65]', 'BB':'[33:34, 32:65]'}
df = my_df.iloc[cr_dict['AA']]
the results I want would be similar to
df = my_df.iloc[42:43,32:65]
I know I could change the dictionary and use the following but it looks convoluted and not as easy to read– is there a better way?
Code
cr_dict= {'AA':[42,43,32,65], 'BB':'[33,34, 32,65]'}
df = my_df.iloc[cr_dict['AA'][0]: cr_dict['AA'][0], cr_dict['AA'][0]: cr_dict['AA'][0]]

Define your dictionaries slightly differently.
cr_dict= {'AA':[42,43]+list(range(32,65)),
'BB':[33,34]+list(range(32,65))}
Then you can slice your DataFrame like so:
>>> my_df.iloc[cr_dict["AA"], cr_dict["BB"]].sort_index()

Python: How to hide index when executing

It's probably something very stupid but I can't find a solution to not print the indexes when executing the code.My code goes:
Reading the excel file and choosing a specific component
df= pd.read_excel('Components.xlsx')
component_name = 'Name'
Forcing the index to be a certain column
df = df.set_index(['TECHNICAL DATA'])
Selecting data in a cell with df.loc
component_lifetime=df.loc[['Life time of Full unit'],component_name]
print(componet_lifetime)
What I get is:
TECHNICAL DATA
Life time of Full unit 20
Is it possible to hide all the index data and only print 20? Thank you ^^

Use pd.DataFrame.at for scalar access by label:
res = df.at['Life time of Full unit', 'Name']
A short guide to indexing:
Use iat / at for scalar access / setting by integer position or label respectively.
Use iloc / loc for non-scalar access / setting by integer position or label respectively.
You can also extract the NumPy array via values, but this is rarely necessary.

Getting a python panda data frame's row name as a string

Having found the maximum value in a panda data frame column, I am just trying to get the equivalent row name as a string.
Here's my code:
df[df['ColumnName'] == df['ColumnName'].max()].index
Which returns me an answer:
Index(['RowName'], dtype='object')
How do I just get RowName back?
(stretch question - why does .idmax() fry in the formulation df['Colname'].idmax? And, yes, I have tried it as .idmax() and also appended it to df.loc[:,'ColName'] etc.)

Just use integer indexing:
df[df['ColumnName'] == df['ColumnName'].max()].index[0]
Here [0] extracts the first element. Note your criterion may yield multiple indices.

Python: How does converters work in genfromtxt() function?

I am new to Python, I have a following example that I don't understand
The following is a csv file with some data
%%writefile wood.csv
item,material,number
100,oak,33
110,maple,14
120,oak,7
145,birch,3
Then, the example tries to define a function to convert those trees name above to integers.
tree_to_int = dict(oak = 1,
maple=2,
birch=3)
def convert(s):
return tree_to_int.get(s, 0)
The first question is why is there a "0" after "s"? I removed that "0" and get same result.
The last step is to read those data by numpy.array
data = np.genfromtxt('wood.csv',
delimiter=',',
dtype=np.int,
names=True,
converters={1:convert}
)
I was wondering for the converters argument, what does {1:convert} exact mean? Especially what does number 1 mean in this case?

For the second question, according to the documentation (https://docs.scipy.org/doc/numpy/reference/generated/numpy.genfromtxt.html), {1:convert} is a dictionary whose keys are column numbers (where the first column is column 0) and whose values are functions that convert the entries in that column.
So in this code, the 1 indicates column one of the csv file, the one with the names of the trees. Including this argument causes numpy to use the convert function to replace the tree names with their corresponding numbers in data.

Tricky str value replacement within PANDAS DataFrame

Problem Overview:
I am attempting to clean stock data loaded from CSV file into Pandas DataFrame. The indexing operation I perform works. If I call print, I can see the values I want are being pulled from the frame. However, when I try to replace the values, as shown in the screenshot, PANDAS ignores my request. Ultimately, I'm just trying to extract a value out of one column and move it to another. The PANDAS documentation suggests using the .replace() method, but that doesn't seem to be working with the operation I'm trying to perform.
Here's a pic of the code and data before and after code is run.
And the for loop (as referenced in the pic):
for i, j in zip(all_exchanges['MarketCap'], all_exchanges['MarketCapSym']):
if 'M' in i: j = j.replace('n/a','M')
elif 'B' in i: j = j.replace('n/a','M')

The problem is that j is a string, thus immutable.
You're replacing data, but not in the original dataset.
You have to do it another way, less elegant, without zip (I simplified your test BTW since it did the same on both conditions):
aem = all_exchanges['MarketCap']
aems = all_exchanges['MarketCapSym']
for i in range(min(len(aem),len(aems)): # like zip: shortest of both
if 'M' in aem[i] or 'B' in aem[i]:
aems[i] = aems[i].replace('n/a','M')
now you're replacing in the original dataset.

If both columns are in the same dataframe, all_exchanges, iterate over the rows.
for i, row in enumerate ( all_exchanges ):
# get whatever you want from row
# using the index you should be able to set a value
all_exchanges.loc[i, 'columnname'] = xyz
That should be the syntax of I remember ;)

Here is quite exhaustive tutorial on missing values and pandas. I suggest using fillna():
df['MarketCap'].fillna('M', inplace=True)
df['MarketCapSym'].fillna('M', inplace=True)

Avoid iterating if you can. As already pointed out, you're not modifying the original data. Index on the MarketCap column and perform the replace as follows.
# overwrites any data in the MarketCapSym column
all_exchanges.loc[(all_exchanges['MarketCap'].str.contains('M|B'),
'MarketCapSym'] = 'M'
# only replaces 'n/a'
all_exchanges.loc[(all_exchanges['MarketCap'].str.contains('M|B'),
'MarketCapSym'].replace({'n/a', 'M'}, inplace=True)

Thanks to all who posted. After thinking about your solutions and the problem a bit longer, I realized there might be a different approach. Instead of initializing a MarketCapSym column with 'n/a', I instead created that column as a copy of MarketCap and then extracted anything that wasn't an "M" or "B".
I was able to get the solution down to one line:
all_exchanges['MarketCapSymbol'] = [ re.sub('[$.0-9]', '', i) for i in all_exchanges.loc[:,'MarketCap'] ]
A breakdown of the solution is as follows:
all_exchanges['MarketCapSymbol'] = - Make a new column on the DataFrame called 'MarketCapSymbol.
all_exchanges.loc[:,'MarketCap'] - Initialize the values in the new column to those in 'MarketCap'.
re.sub('[$.0-9]', '', i) for i in - Since all I want is the 'M' or 'B', apply re.sub() on each element, extracting [$.0-9] and leaving only the M|B.
Using a list comprehension this way seemed a bit more natural / readable to me in my limited experience with PANDAS. Let me know what you think!