[英]How do you apply a function on a dataframe column using data from previous rows?
我有一个 Dataframe ,它有三列:带有一些值的nums
, b
始终为1
或0
, result
列当前在除第一行之外的任何地方都为零(因为我们必须有一个初始值才能使用)。 dataframe 看起来像这样:
nums b result
0 20.0 1 20.0
1 22.0 0 0
2 30.0 1 0
3 29.1 1 0
4 20.0 0 0
...
我想从第二行开始对 dataframe 中的每一行进行 go ,进行一些计算并将结果存储在result
列中。 由于我正在处理大文件,因此我需要一种方法来快速执行此操作,这就是我想要类似apply
的原因。
我要做的计算是从前一行获取nums
和result
中的值,如果在当前行中b
col 为0
,那么我想要(例如)添加前一行的num
和result
. 例如,如果该行中的b
为1
,我想减去它们。
我尝试使用apply
但我无法访问上一行,遗憾的是,如果我设法访问上一行,dataframe 直到最后都不会更新结果列。
我也尝试过使用这样的循环,但是对于我正在使用的大型文件来说它太慢了:
for i in range(1, len(df.index)):
row = df.index[i]
new_row = df.index[i - 1] # get index of previous row for "nums" and "result"
df.loc[row, 'result'] = some_calc_func(prev_result=df.loc[new_row, 'result'], prev_num=df.loc[new_row, 'nums'], \
current_b=df.loc[row, 'b'])
some_calc_func
看起来像这样(只是一个一般示例):
def some_calc_func(prev_result, prev_num, current_b):
if current_b == 1:
return prev_result * prev_num / 2
else:
return prev_num + 17
请回答some_calc_func
如果您想保留 function some_calc_func
并且不使用另一个库,则不应尝试在每次迭代时访问每个元素,您可以在 nums 和 b 列上使用zip
并在两者之间转换,因为您尝试访问前一个行并将每次迭代的 prev_res 保存在 memory 中。 此外, append
到一个列表而不是 dataframe,并在循环之后将列表分配给列。
prev_res = df.loc[0, 'result'] #get first result
l_res = [prev_res] #initialize the list of results
# loop with zip to get both values at same time,
# use loc to start b at second row but not num
for prev_num, curren_b in zip(df['nums'], df.loc[1:, 'b']):
# use your function to calculate the new prev_res
prev_res = some_calc_func (prev_res, prev_num, curren_b)
# add to the list of results
l_res.append(prev_res)
# assign to the column
df['result'] = l_res
print (df) #same result than with your method
nums b result
0 20.0 1 20.0
1 22.0 0 37.0
2 30.0 1 407.0
3 29.1 1 6105.0
4 20.0 0 46.1
现在有了 5000 行的 dataframe df,我得到:
%%timeit
prev_res = df.loc[0, 'result']
l_res = [prev_res]
for prev_num, curren_b in zip(df['nums'], df.loc[1:, 'b']):
prev_res = some_calc_func (prev_res, prev_num, curren_b)
l_res.append(prev_res)
df['result'] = l_res
# 4.42 ms ± 695 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
使用您的原始解决方案,速度慢了约 750 倍
%%timeit
for i in range(1, len(df.index)):
row = df.index[i]
new_row = df.index[i - 1] # get index of previous row for "nums" and "result"
df.loc[row, 'result'] = some_calc_func(prev_result=df.loc[new_row, 'result'], prev_num=df.loc[new_row, 'nums'], \
current_b=df.loc[row, 'b'])
#3.25 s ± 392 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
如果 function some_calc_func
可以很容易地与 Numba 装饰器一起使用,则使用另一个名为numba
的库进行编辑。
from numba import jit
# decorate your function
@jit
def some_calc_func(prev_result, prev_num, current_b):
if current_b == 1:
return prev_result * prev_num / 2
else:
return prev_num + 17
# create a function to do your job
# numba likes numpy arrays
@jit
def with_numba(prev_res, arr_nums, arr_b):
# array for results and initialize
arr_res = np.zeros_like(arr_nums)
arr_res[0] = prev_res
# loop on the length of arr_b
for i in range(len(arr_b)):
#do the calculation and set the value in result array
prev_res = some_calc_func (prev_res, arr_nums[i], arr_b[i])
arr_res[i+1] = prev_res
return arr_res
最后,像这样称呼它
df['result'] = with_numba(df.loc[0, 'result'],
df['nums'].to_numpy(),
df.loc[1:, 'b'].to_numpy())
通过 timeit,我比使用 zip 的方法快了约 9 倍,并且速度会随着尺寸的增加而增加
%timeit df['result'] = with_numba(df.loc[0, 'result'],
df['nums'].to_numpy(),
df.loc[1:, 'b'].to_numpy())
# 526 µs ± 45.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
注意使用 Numba 可能会出现问题,具体取决于您的实际some_calc_func
国际大学联盟:
>>> df['result'] = (df[df.result.eq(0)].b.replace({0: 1, 1: -1}) * df.nums
).fillna(df.result).cumsum()
>>> df
nums b result
0 20.0 1 20.0
1 22.0 0 42.0
2 30.0 1 12.0
3 29.1 1 -17.1
4 20.0 0 2.9
解释:
# replace 0 with 1 and 1 with -1 in column `b` for rows where result==0
>>> df[df.result.eq(0)].b.replace({0: 1, 1: -1})
1 1
2 -1
3 -1
4 1
Name: b, dtype: int64
# multiply with nums
>>> (df[df.result.eq(0)].b.replace({0: 1, 1: -1}) * df.nums)
0 NaN
1 22.0
2 -30.0
3 -29.1
4 20.0
dtype: float64
# fill the 'NaN' with the corresponding value from df.result (which is 20 here)
>>> (df[df.result.eq(0)].b.replace({0: 1, 1: -1}) * df.nums).fillna(df.result)
0 20.0
1 22.0
2 -30.0
3 -29.1
4 20.0
dtype: float64
# take the cumulative sum (cumsum)
>>> (df[df.result.eq(0)].b.replace({0: 1, 1: -1}) * df.nums).fillna(df.result).cumsum()
0 20.0
1 42.0
2 12.0
3 -17.1
4 2.9
dtype: float64
根据您在评论中的要求,我想不出没有循环的方法:
c1, c2 = 2, 1
l = [df.loc[0, 'result']] # store the first result in a list
# then loop over the series (df.b * df.nums)
for i, val in (df.b * df.nums).iteritems():
if i: # except for 0th index
if val == 0: # (df.b * df.nums) == 0 if df.b == 0
l.append(l[-1]) # append the last result
else: # otherwise apply the rule
t = l[-1] *c2 + val * c1
l.append(t)
>>> l
[20.0, 20.0, 80.0, 138.2, 138.2]
>>> df['result'] = l
nums b result
0 20.0 1 20.0
1 22.0 0 20.0
2 30.0 1 80.0 # [ 20 * 1 + 30 * 2]
3 29.1 1 138.2 # [ 80 * 1 + 29.1 * 2]
4 20.0 0 138.2
似乎足够快,没有测试大样本。
你有 af(...) 申请,但不能因为你需要保留一个 memory (前一个)行。 您可以使用闭包或 class 来执行此操作。 下面是一个 class 实现:
import pandas as pd
class Func():
def __init__(self, value):
self._prev = value
self._init = True
def __call__(self, x):
if self._init:
res = self._prev
self._init = False
elif x.b == 0:
res = x.nums - self._prev
else:
res = x.nums + self._prev
self._prev = res
return res
#df = pd.read_clipboard()
f = Func(20)
df['result'] = df.apply(f, axis=1)
您可以用some_calc_func
正文中的任何内容替换__call__
。
我意识到这就是@Prodipta 的答案,但这种方法使用global
关键字来记住每次迭代apply
的先前结果:
prev_result = 20
def my_calc(row):
global prev_result
i = int(row.name) #the index of the current row
if i==0:
return prev_result
elif row['b'] == 1:
out = prev_result * df.loc[i-1,'nums']/2 #loc to get prev_num
else:
out = df.loc[i-1,'nums'] + 17
prev_result = out
return out
df['result'] = df.apply(my_calc, axis=1)
您的示例数据的结果:
nums b result
0 20.0 1 20.0
1 22.0 0 37.0
2 30.0 1 407.0
3 29.1 1 6105.0
4 20.0 0 46.1
这是@Ben T的答案的速度测试-不是最好的,但不是最差的?
In[0]
df = pd.DataFrame({'nums':np.random.randint(0,100,5000),'b':np.random.choice([0,1],5000)})
prev_result = 20
%%timeit
df['result'] = df.apply(my_calc, axis=1)
Out[0]
117 ms ± 5.67 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
重用你的循环和 some_calc_func
我正在使用您的循环并将其减少到最低限度,如下所示
for i in range(1, len(df)):
df.loc[i, 'result'] = some_calc_func(df.loc[i, 'b'], df.loc[i - 1, 'result'], df.loc[i, 'nums'])
并且some_calc_func
实现如下
def some_calc_func(bval, prev_result, curr_num):
if bval == 0:
return prev_result + curr_num
else:
return prev_result - curr_num
结果如下
nums b result
0 20.0 1 20.0
1 22.0 0 42.0
2 30.0 1 12.0
3 29.1 1 -17.1
4 20.0 0 2.9
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