You need to check all numbers from 2 to n-1 (to sqrt(n) actually, but ok, let it be n). If n is divisible by any of the numbers, it is not prime. If a number is prime, print it. for num in range(2,101): prime = True for i in range(2,num): if (num%i==0): prime = False … Read more
As Mohit Motwani suggested fastest way is to collect data into dictionary then load all into data frame. Below some speed measurements examples: import pandas as pd import numpy as np import time import random end_value = 10000 Measurement for creating a list of dictionaries and at the end load all into data frame start_time … Read more
If I understand correctly you can add query(“Points > 100”): df = pd.DataFrame({‘Points’:[50,20,38,90,0, np.Inf], ‘Player’:[‘a’,’a’,’a’,’s’,’s’,’s’]}) print (df) Player Points 0 a 50.000000 1 a 20.000000 2 a 38.000000 3 s 90.000000 4 s 0.000000 5 s inf points_series = df.query(“Points < inf”).groupby(“Player”).agg({“Points”: “sum”})[‘Points’] print (points_series) a = points_series[points_series > 100] print (a) Player a 108.0 … Read more
Another way is to extract the values of your series and use numpy.stack on them. np.stack(s.values) PS. I’ve run into similar situations often.
I’m using pandas 0.23.3 and Python 3.6, so I can see a real difference in running time only for your second example. But let’s investigate a slightly different version of your second example (so we get2*df[0] out of the way). Here is our baseline on my machine: twice = df[0]*2 mask = df[0] > 0.5 … Read more
This might not be obvious, but pd.Series.isin uses O(1)-look up per element. After an analysis, which proves the above statement, we will use its insights to create a Cython-prototype which can easily beat the fastest out-of-the-box-solution. Let’s assume that the “set” has n elements and the “series” has m elements. The running time is then: … Read more
Just call reset_index(): In [130]: s Out[130]: 0 1 1999-03-31 SOLD_PRICE NaN 1999-06-30 SOLD_PRICE NaN 1999-09-30 SOLD_PRICE NaN 1999-12-31 SOLD_PRICE 3 2000-03-31 SOLD_PRICE 3 Name: 2, dtype: float64 In [131]: s.reset_index() Out[131]: 0 1 2 0 1999-03-31 SOLD_PRICE NaN 1 1999-06-30 SOLD_PRICE NaN 2 1999-09-30 SOLD_PRICE NaN 3 1999-12-31 SOLD_PRICE 3 4 2000-03-31 SOLD_PRICE 3 … Read more
You can use reindex of index of Series created by len with sort_values: print (df.name.str.len()) 0 5 1 2 2 6 3 4 Name: name, dtype: int64 print (df.name.str.len().sort_values()) 1 2 3 4 0 5 2 6 Name: name, dtype: int64 s = df.name.str.len().sort_values().index print (s) Int64Index([1, 3, 0, 2], dtype=”int64″) print (df.reindex(s)) name score … Read more
Well it seems simply zip(s,s.index) works too! For Python-3.x, we need to wrap it with list – list(zip(s,s.index)) To get a tuple of tuples, use tuple() : tuple(zip(s,s.index)). Sample run – In [8]: s Out[8]: a 1 b 2 c 3 dtype: int64 In [9]: list(zip(s,s.index)) Out[9]: [(1, ‘a’), (2, ‘b’), (3, ‘c’)] In [10]: … Read more
I also like itertuples() for row in df.itertuples(): print(row.A) print(row.Index) since row is a named tuples, if you meant to access values on each row this should be MUCH faster speed run : df = pd.DataFrame([x for x in range(1000*1000)], columns=[‘A’]) st=time.time() for index, row in df.iterrows(): row.A print(time.time()-st) 45.05799984931946 st=time.time() for row in df.itertuples(): … Read more