Check type of your numbers before converting them to strings. It seems that they are floats, rather than integers. If this is the case, convert your numbers to integers:

df = pd.DataFrame([123.0, 456.0])
df = df.apply(int, axis=1)

0    123
1    456

Then, convert it into strings:

df = df.apply(str)
print(df.iloc[1])

'456'

Since pandas 0.17.1 you can set the displayed numerical precision by modifying the style of the particular data frame rather than setting the global option:

import pandas as pd
import numpy as np

np.random.seed(24)
df = pd.DataFrame(np.random.randn(5, 3), columns=list('ABC'))
df 

df.style.set_precision(2)

It is also possible to apply column specific styles

df.style.format({
    'A': '{:,.1f}'.format,
    'B': '{:,.3f}'.format,
})

Your question has nothing to do with Spark or PySpark. It's related to Pandas.

This is because Pandas interpret and infer columns' data type automatically. Since all the values of your column are numeric, Pandas will consider it as float data type.

To avoid this, pandas.ExcelFile.parse method accepts an argument called converters, you could use this to tell Pandas the specific column data type by:

# if you want one specific column as string
df = pd.concat([filepath_pd.parse(name, converters={'column_name': str}) for name in names])

OR

# if you want all columns as string
# and you have multi sheets and they do not have same columns
# this merge all sheets into one dataframe
def get_converters(excel_file, sheet_name, dt_cols):
    cols = excel_file.parse(sheet_name).columns
    converters = {col: str for col in cols if col not in dt_cols}
    for col in dt_cols:
        converters[col] = pd.to_datetime
    return converters

df = pd.concat([filepath_pd.parse(name, converters=get_converters(filepath_pd, name, ['date_column'])) for name in names]).reset_index(drop=True)

OR

# if you want all columns as string
# and all your sheets have same columns
cols = filepath_pd.parse().columns
dt_cols = ['date_column']
converters = {col: str for col in cols if col not in dt_cols}
for col in dt_cols:
    converters[col] = pd.to_datetime
df = pd.concat([filepath_pd.parse(name, converters=converters) for name in names]).reset_index(drop=True)

You can solve this by setting the pandas option, precision to 0.

import pandas as pd
df = pd.DataFrame(data={"col": [24.00, 2.00, 3.00]})
print(df)
    col
0  24.0
1   2.0
2   3.0
pd.set_option('precision',0)
print(df)
   col
0   24
1    2
2    3

Here is hint for you ,

In case of Valid number ,

a="2012.0"
try:
    a=float(a)
    a=int(a)
    print a

except:
    print a

Output:

2012

In case of String like "Dance with Me"

a="Dance with Me"
try:
    a=float(a)
    a=int(a)
    print a

except:
    print a

Output:

Dance with Me

In Pandas/NumPy, integers are not allowed to take NaN values, and arrays/series (including dataframe columns) are homogeneous in their datatype --- so having a column of integers where some entries are None/np.nan is downright impossible.

EDIT:data.phone.astype('object') should do the trick; in this case, Pandas treats your column as a series of generic Python objects, rather than a specific datatype (e.g. str/float/int), at the cost of performance if you intend to run any heavy computations with this data (probably not in your case).

Assuming you want to keep those NaN entries, your approach of converting to strings is a valid possibility:

data.phone.astype(str).str.split('.', expand = True)[0]

should give you what you're looking for (there are alternative string methods you can use, such as .replace or .extract, but .split seems the most straightforward in this case).

Alternatively, if you are only interested in the display of floats (unlikely I'd suppose), you can do pd.set_option('display.float_format','{:.0f}'.format), which doesn't actually affect your data.

You can coerce the datatype to int, Just a note in case you have nans in your data, the conversion to int doesn't work as they have float data type, so regex solution might be better.

df['Net Sales'] = df['Net Sales'].astype('int') 

or in case of regex:

df['Net Sales'] = df['Net Sales'].astype('str').replace(r'\.\d+$', '', regex=True).astype('int')

Example:

import pandas as pd

df = pd.DataFrame({"Net Sales" : [1.5, 2.5]})

df['Net Sales'] = df['Net Sales'].astype('int')

df['Net Sales'] = df['Net Sales'].astype('str').replace(r'\.\d+$', '', regex=True).astype('int')

Output:

#   Net Sales
#0  1
#1  2

If - as your comment suggests - the data just all needs to be multiplied by 100...

df['columnName'] = df['columnName'].apply(lambda x: x*100)

You can do it like so:

df.Name = df.Name.str.replace('\d+', '')

To play and explore, check the online Regular expression demo here: https://regex101.com/r/Y6gJny/2

Whatever is matched by the pattern \d+ i.e 1 or more digits, will be replaced by empty string.