如何从字典键的前面删除多余的空间？

Question

I have a dictionary. There are some space infant of keys. How I can remove all of these space and keep only one space infant of keys? I made the dictionary with the following code:

def get_JobAbbreviation_data():
    data=pd.read_csv('A.csv')
    data=data.replace(u'\xa0', u'', regex=True)
    data.dropna(inplace=True)
    Title = data['FIRST'].str.lower()+' '
    Abbr = data['1ST'].str.lower()+ ' '
    JobAbbreviation=dict(zip(Abbr, Title))
    return JobAbbreviation

Answer 1

This breaks down to "How to remove all (but one) spaces at the start of a string"?

Removing spaces is easy:

In [1]: "    abcd".lstrip()
Out[1]: 'abcd'

Then, just add a space, so... " " + mystring.lstrip() .

That's the building block. The next part of the question is: "How to apply this to all the keys?":

Abbr = [" " + s.lstrip() for s in Abbr]

Then build your dictionary.

Answer 2

I assume by "infant" you actually mean "in front". If you have a dictionary that looks as follows:

dct = {"   Three Spaces": 3, "    Four Spaces": 4, " One Space": 1, "Zero Spaces": 0}

And you want this exact same dictionary where all the keys have only 1 white space at the start, you can do it as follows:

dctKeys = list(map((lambda x: " " + x.strip()) , [*dct.keys()]))
finDct = dict(zip(dctKeys, dct.values()))
finDct

This output will be the following:

{' Four Spaces': 4, ' One Space': 1, ' Three Spaces': 3, ' Zero Spaces': 0}