I have a dataframe where some index number are missing how do I cut dataframe previous to that missing index number

tez 0
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[enter image description here][1]

Index number 72 is missing from original dataframe which is shown in image. I want to cut dataframe like [0:71,:] with condition like when index sequence breaks then dataframe automatically cuts from previous index value.

dataframe pandas python-3.x python
3 Answers

Compare shifted values of index subtracted by original values if greater like 1 with invert ordering by [::-1] and Series.cummax, last filter in boolean indexing:

df = pd.DataFrame({'a': range(3,13)}).drop(3)
print (df)
    a
0   3
1   4
2   5
4   7
5   8
6   9
7  10
8  11
9  12

df = df[df.index.to_series().shift(-1, fill_value=0).sub(df.index).gt(1)[::-1].cummax()]
print (df)
   a
0  3
1  4
2  5
jezrael0

i came to this:

df = pd.DataFrame({'col':[1,2,3,4,5,6,7,8,9]}, index=[-1,0,1,2,3,4,5,7,8])

ind = next((i for i in range(len(df)-1) if df.index[i]+1!=df.index[i+1]),len(df))+1

>>> df.iloc[:ind]
'''
    col
-1    1
 0    2
 1    3
 2    4
 3    5
 4    6
 5    7
SergFSM0

With numpy, get the values that are equal to a normal range starting from the first index, up to the first mismatch (excluded):

df[np.minimum.accumulate(df.index==np.arange(df.index[0], df.index[0]+len(df)))]

Example:

    col
-1    1
 0    2
 1    3
 3    4
 4    5

output:

    col
-1    1
 0    2
 1    3
mozway0
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