I have the following example of dataframe.
c1 c2 0 1 a 1 2 b 2 3 c 3 4 d 4 5 e Given a template c1 = [3, 2, 5, 4, 1], I want to change the order of the rows based on the new order of column c1, so it will look like:
c1 c2 0 3 c 1 2 b 2 5 e 3 4 d 4 1 a I found the following thread, but the shuffle is random. Cmmiw.
If values are unique in list and also in c1 column use reindex:
df = df.set_index('c1').reindex(c1).reset_index() print (df) c1 c2 0 3 c 1 2 b 2 5 e 3 4 d 4 1 a General solution working with duplicates in list and also in column:
c1 = [3, 2, 5, 4, 1, 3, 2, 3] #create df from list list_df = pd.DataFrame({'c1':c1}) print (list_df) c1 0 3 1 2 2 5 3 4 4 1 5 3 6 2 7 3 #helper column for count duplicates values df['g'] = df.groupby('c1').cumcount() list_df['g'] = list_df.groupby('c1').cumcount() #merge together, create index from column and remove g column df = list_df.merge(df).drop('g', axis=1) print (df) c1 c2 0 3 c 1 2 b 2 5 e 3 4 d 4 1 a 5 3 c 
3 and 2 in list, then reindex raise error. So is necessary count each value by cumcount and then merge - first dupe of list with first dupe of column.mergeYou can create a dataframe with the column specified in the wanted order then merge.
One advantage of this approach is that it gracefully handles duplicates in either df.c1 or the list c1. If duplicates not wanted then care must be taken to handle them prior to reordering.
d1 = pd.DataFrame({'c1': c1}) d1.merge(df) c1 c2 0 3 c 1 2 b 2 5 e 3 4 d 4 1 a searchsortedThis is less robust but will work if df.c1 is:
df.iloc[df.c1.searchsorted(c1)] c1 c2 2 3 c 1 2 b 4 5 e 3 4 d 0 1 a 