从不同行的列中为 pandas DataFrame 列分配值的最佳方法是什么？

Question

I need to iterate over a DataFrame indexed by UNIX timestamp, and in one column, assign a value from another column in a different row at a specific index time in the future. This is what I'm currently doing:

df = pd.DataFrame([
    [1523937600, 100.0, 0.0], 
    [1523937660, 120.0, 0.0], 
    [1523937720, 110.0, 0.0],
    [1523937780, 90.0, 0.0],
    [1523937840, 99.0, 0.0]], 
    columns=['time', 'value', 'target'])
df.set_index('time', inplace=True)

skip = 2  # mins skip-ahead
for i in range(0, df.shape[0]-1):       
    t = df.index[i] + (60*skip)
    try:
        df.iloc[i].target = df.loc[t].value
    except KeyError:
        df.iloc[i].target = 0.0

Output:

            value  target
time                     
1523937600  100.0   110.0
1523937660  120.0    90.0
1523937720  110.0    99.0
1523937780   90.0     0.0
1523937840   99.0     0.0

This works, but I am dealing with datasets containing millions of rows and it takes an extremely long time. Is there a more optimal way to do this?

EDIT: Added example input/output. Note, it is important that I obtain the value from the row with the calculated index time rather than just look ahead n rows, as there could be gaps in the times, or additional times in between.

Answer 1

In this case you should keep time as a column as well as index. Hope this helps:

df = pd.DataFrame([ 
    [1523937600, 100.0, 0.0], 
    [1523937660, 120.0, 0.0], 
    [1523937720, 110.0, 0.0],
    [1523937780, 90.0, 0.0],
    [1523937840, 99.0, 0.0]], 
    columns=['time', 'value', 'target'])
df.index = df['time']

df['target'] = df['time'].apply(lambda x: df.loc[x+(skip*60)].value if x+(skip*60) in df.index.values  else 0.0)