Select Pandas rows based on list index

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Question :

Select Pandas rows based on list index

I have a dataframe df :

   20060930  10.103       NaN     10.103   7.981
   20061231  15.915       NaN     15.915  12.686
   20070331   3.196       NaN      3.196   2.710
   20070630   7.907       NaN      7.907   6.459

Then I want to select rows with certain sequence numbers which indicated in a list, suppose here is [1,3], then left:

   20061231  15.915       NaN     15.915  12.686
   20070630   7.907       NaN      7.907   6.459

How or what function can do that ?

Answer #1:

ind_list = [1, 3]

should do the trick!
When I index with data frames I always use the .ix() method. Its so much easier and more flexible…

This is no longer the accepted method for indexing. The ix method is deprecated. Use .iloc for integer based indexing and .loc for label based indexing.

Answered By: Woody Pride

Answer #2:

you can also use iloc:


This will not work if the indexes in your dataframe do not correspond to the order of the rows due to prior computations. In that case use:


… as suggested in other responses.

Answered By: yemu

Answer #3:

Another way (although it is a longer code) but it is faster than the above codes. Check it using %timeit function:


PS: You figure out the reason

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Answered By: Amruth Lakkavaram

Answer #4:

For large datasets, it is memory efficient to read only selected rows via the skiprows parameter.


pred = lambda x: x not in [1, 3]
pd.read_csv("data.csv", skiprows=pred, index_col=0, names=...)

This will now return a DataFrame from a file that skips all rows except 1 and 3.


From the docs:

skiprows : list-like or integer or callable, default None

If callable, the callable function will be evaluated against the row indices, returning True if the row should be skipped and False otherwise. An example of a valid callable argument would be lambda x: x in [0, 2]

This feature works in version pandas 0.20.0+. See also the corresponding issue and a related post.

Answered By: pylang

Answer #5:

There are many ways of solving this problem, and the ones listed above are the most commonly used ways of achieving the solution. I want to add two more ways, just in case someone is looking for an alternative.

index_list = [1,3]



df.query('index in @index_list')
Answered By: Loochie

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