Develop Reference

How to change 0.80 to 0.8 / 1.0 to 1 in dataframe?

I have a dataframe looks like this.
df = pd.DataFrame([0.70,1.0,0.75,0.0,5.0], columns=['pitch'])
I want to convert it into
df = pd.DataFrame([0.7,1,0.75,0,5], columns=['pitch'])
If i convert the float to int, 0.7 will be 0 .
How to solve this problem, thanks!!

astype()
e round
can use the astype(int) function in conjunction with the round() function on the dataframe. This will round each value in the 'pitch' column to the desired number of decimal places and then convert those rounded values to integers.
df['pitch'] = df['pitch'].round(1).astype(int)
or using round(0) to remove decimal places completely
df['pitch'] = df['pitch'].round(0).astype(int)
Be careful using the astype(int) function to convert to int, as it can round values, not just remove decimal places.

You can use the round() function to round the decimal values to the desired number of decimal places before converting them to integers. For example, you can use round(df['pitch'], 1) to round the values in the 'pitch' column to one decimal place, then use astype(int) to convert them to integers.
df['pitch'] = round(df['pitch'], 1).astype(int)

In pandas, how to convert the result of dividing two columns from a decimal to a percentage?

When using pandas, how do I convert a decimal to a percentage when the result obtained by dividing two columns as another column?
for example :
df_income_structure['C'] = (df_income_structure['A']/df_income_structure['B'])
If the value of df_income_structure['C'] is a decimal, how to convert it to a percentage ?

Format it like this:
df_income_structure.style.format({'C': '{:,.2%}'.format})
Change the number depending on how many decimal places you'd like.

Use basic pandas operators
For example if we have a dataframe with columns name like column1, column2,column3 , ... so we can :
Columns = [column1, column2,column3, ....] .
df[Columns] = df[Columns].div(df[Columns].sum(axis=1), axis=0).multiply(100)
(df[Columns].sum(axis=1). axis=1 makes the summation for rows.
Divide the dataframe by (df[Columns].div(df[Columns].sum(axis=1), axis=0). axis=0 is for devision of columns.
multiply the results by 100 for percentages of 100.
I hope this answer has solved your problem.
Good luck

convert float64 to int (excel to pandas)

I have imported excel file into python pandas. but when I display customer numbers I get in float64 format i.e
7.500505e+09 , 7.503004e+09
how do convert the column containing these numbers

int(yourVariable) will cast your float64 to a integer number.
Is this what you are looking for?

You can use pandas DataFrames style.format function to apply a formatting to a column like in https://towardsdatascience.com/style-pandas-dataframe-like-a-master-6b02bf6468b0. If you want to round the numbers to i.e. 2 decimal places follow the accepted answer in Python float to Decimal conversion. Conversion to an int reduces accuracy a bit too much i think
For the conecpt binary floats in general (what float64 is) see Python - round a float to 2 digits

Please use pd.set_option to control the precision's to be viewed
>>> import pandas as pd
>>> a = pd.DataFrame({'sam':[7.500505e+09]})
>>> pd.set_option('float_format', '{:f}'.format)
>>> a
sam
0 7500505000.000000
>>>

Pandas Dataframe Float Precision

I am trying to alter my dataframe with the following line of code:
df = df[df['P'] <= cutoff]
However, if for example I set cutoff to be 0.1, numbers such as 0.100496 make it through the filter.
My suspicion is that my initial dataframe has entries in scientific notation and float format as well. Could this be affecting the rounding and precision? Is there a potential workaround to this issue.
Thank you in advance.
EDIT: I am reading from a file. Here is a sample of the total data.
2.29E-98
1.81E-42
2.19E-35
3.35E-30
0.0313755
0.0313817
0.03139
0.0313991
0.0314062
0.1003476
0.1003483
0.1003487
0.1003521
0.100496

Floating point comparison isn't perfect. For example
>>> 0.10000000000000000000000000000000000001 <= 0.1
True
Have a look at numpy.isclose. It allows you to compare floats and set a tolerance for the comparison.
Similar question here

Selecting rows by value in a floating point column in pandas

I import a csv data file into a pandas DataFrame df with pd.read_csv. The text file contains a column with strings like these:
y
0.001
0.0003
0.0001
3e-05
1e-05
1e-06
If I print the DataFrame, pandas outputs the decimal representation of these values with 6 digits after the comma, and everything looks good.
When I try to select rows by value, like here:
df[df['y'] == value],
by typing the corresponding decimal representation of value, pandas correctly matches certain values (example: rows 0, 2, 4) but does not match others (rows 1, 3, 5). This is of course due to the fact that those rows values do not have a perfect representation in base two.
I was able to workaround this problem is this way:
df[abs(df['y']/value-1) <= 0.0001]
but it seems somewhat awkward. What I'm wondering is: numpy already has a method, .isclose, that is specifically for this purpose.
Is there a way to use .isclose in a case like this? Or a more direct solution in pandas?

Yes, you can use numpy's isclose
df[np.isclose(df['y'], value)]