pandas 使用股票报价识别熊猫数据框中的价格波动/趋势
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Identifying price swings/trends in pandas dataframe with stock quotes
提问by Ovidiu Susan
I have a pandas Dataframe with DatetimeIndex and ohlcv stock quotes columns. I would like to extract price swings/trends that meet a certain threshold: up swings/trends/moves bigger than 0.3$ and down swings/trends/moves that go beyond -0.3$.
我有一个带有 DatetimeIndex 和 ohlcv 股票报价列的Pandas数据框。我想提取满足特定阈值的价格波动/趋势：大于 0.3 美元的向上波动/趋势/移动和超过 -0.3 美元的向下波动/趋势/移动。
df[:10]
close high low open volume
2014-05-09 09:30:00-04:00 187.5600 187.73 187.54 187.700 1922600
2014-05-09 09:31:00-04:00 187.4900 187.56 187.42 187.550 534400
2014-05-09 09:32:00-04:00 187.4200 187.51 187.35 187.490 224800
2014-05-09 09:33:00-04:00 187.5500 187.58 187.39 187.400 303700
2014-05-09 09:34:00-04:00 187.6700 187.67 187.53 187.560 438200
2014-05-09 09:35:00-04:00 187.6000 187.71 187.56 187.680 296400
2014-05-09 09:36:00-04:00 187.4100 187.67 187.38 187.600 329900
2014-05-09 09:37:00-04:00 187.3100 187.44 187.28 187.400 404000
2014-05-09 09:38:00-04:00 187.2600 187.37 187.26 187.300 912800
2014-05-09 09:39:00-04:00 187.2200 187.28 187.12 187.250 607700
After studying pandas documentation it looked like the Dataframe.apply() would be the approach, but I got stuck in building the function(s). As my coding abilities are limited overall I need a little help please.
在研究了 Pandas 文档后，看起来 Dataframe.apply() 将是一种方法，但我陷入了构建函数的困境。由于我的编码能力总体有限，我需要一些帮助。
global row_nr
row_nr = 1
def extract_swings()
if row_nr == 1 : pivot = row.open ; row_nr += 1
else : if (row.high-pivot) >= 0.3 : ????
... ????
df['swings'] = df.apply(extract_swings, axis=1)
The result should be this:
结果应该是这样的：
df['swings'][:10]
2014-05-09 09:30:00-04:00 NaN
2014-05-09 09:31:00-04:00 NaN
2014-05-09 09:32:00-04:00 -0.35
2014-05-09 09:33:00-04:00 NaN
2014-05-09 09:34:00-04:00 NaN
2014-05-09 09:35:00-04:00 0.36
2014-05-09 09:36:00-04:00 NaN
2014-05-09 09:37:00-04:00 NaN
2014-05-09 09:38:00-04:00 NaN
2014-05-09 09:39:00-04:00 -0.59
UPDATE: To avoid any confusion here is how the requested function should go through the dataframe:
更新：为了避免混淆，这里是请求的函数应该如何通过数据帧：
close high low open volume
2014-05-09 09:30:00-04:00 187.5600 187.73 187.54 187.700 1922600
# this is the first line, first minute and we well take row.open 187.70 as \
# the starting point or first pivot
2014-05-09 09:31:00-04:00 187.4900 187.56 187.42 187.550 534400
# next minute we check if either (row.high - pivot) >= 0.3 or \
# (row.low-pivot) <= -0.3. Neither is true so nothing to do here.
2014-05-09 09:32:00-04:00 187.4200 187.51 187.35 187.490 224800
# next minute same check ... we see that row.low-pivot = -0.35. \
# We consider 187.35 a second pivot and the diff value -0.35 a first trend down
2014-05-09 09:33:00-04:00 187.5500 187.58 187.39 187.400 303700
# next minute we check if the identified trend/swing down goes further \
# down by having a row.low lower than previous row.low. If we would \
# have found here a new lower row.low that would be the second pivot \
# and we would forget about 187.35 as being a pivot ... and so on. \
# We don't see that on this row, instead we see prices are higher than \
# previous row, so we start checking the diff for a potential up trend \
# starting from second pivot 187.35. As long as we do not encounter a \
# higher high with over 0.3 above last pivot we are still within the identified down trend.
2014-05-09 09:34:00-04:00 187.6700 187.67 187.53 187.560 438200
# we don't see a lower low to reconsider the second pivot neither \
# a (row.high- second_pivot) >= 0.3
2014-05-09 09:35:00-04:00 187.6000 187.71 187.56 187.680 296400
# here we see (row.high- second_pivot) = 0.36. We consider 187.71 as \
# a third_pivot and the diff value 0.36 as an up trend (from second pivot to here)
2014-05-09 09:36:00-04:00 187.4100 187.67 187.38 187.600 329900
# next minute we check if the identified trend/swing up goes further up \
# by having a row.high higher than third pivot. If we would have found here \
# a new higher row.high that would be the third pivot and we would forget \
# about 187.71 as being a pivot ... and so on. We don't see that on this row,\
# instead we see prices are lower than previous row, so we start \
# checking the diff for a potential down trend starting from third \
# pivot 187.71. As long as we do not encounter a lower low with \
# over 0.3 below last pivot we are still within the identified up trend.
2014-05-09 09:37:00-04:00 187.3100 187.44 187.28 187.400 404000
# we find here a (row.low - third_pivot) = 0.43 so we have identified \
# a new down trend starting from third pivot and now we have a potential\
# fourth pivot 187.28
2014-05-09 09:38:00-04:00 187.2600 187.37 187.26 187.300 912800
# we find here a lower low so we don't consider 187.28 the fourth \
# pivot anymore but this lower low 187.26
2014-05-09 09:39:00-04:00 187.2200 187.28 187.12 187.250 607700
# we find here a lower low so we don't consider 187.26 the fourth pivot anymore \
# but this lower low 187.12. Being this the lowest low we consider this one \
# to be the fourth pivot and the diff 187.12-187.71=-0.59 as a downtrend with that value
回答by Pawel Kozela
It's a bit tricky since you cannot mark a point as pivot until you find the next potential pivot (ie if you are in an upward trend, you can't say it's done until you find a low sufficiently low).
这有点棘手，因为在找到下一个潜在支点之前，您无法将一个点标记为支点（即，如果您处于上升趋势中，则在找到足够低的低点之前不能说它已经完成）。
This code does the trick - I've put your data in the tmpData.txt file for convenience, and get the desired result. Please check
这段代码可以解决问题 - 为方便起见，我已将您的数据放在 tmpData.txt 文件中，并获得所需的结果。请检查
def get_pivots():
data = pd.DataFrame.from_csv('tmpData.txt')
data['swings'] = np.nan
pivot = data.irow(0).open
last_pivot_id = 0
up_down = 0
diff = .3
for i in range(0, len(data)):
row = data.irow(i)
# We don't have a trend yet
if up_down == 0:
if row.low < pivot - diff:
data.ix[i, 'swings'] = row.low - pivot
pivot, last_pivot_id = row.low, i
up_down = -1
elif row.high > pivot + diff:
data.ix[i, 'swings'] = row.high - pivot
pivot, last_pivot_id = row.high, i
up_down = 1
# Current trend is up
elif up_down == 1:
# If got higher than last pivot, update the swing
if row.high > pivot:
# Remove the last pivot, as it wasn't a real one
data.ix[i, 'swings'] = data.ix[last_pivot_id, 'swings'] + (row.high - data.ix[last_pivot_id, 'high'])
data.ix[last_pivot_id, 'swings'] = np.nan
pivot, last_pivot_id = row.high, i
elif row.low < pivot - diff:
data.ix[i, 'swings'] = row.low - pivot
pivot, last_pivot_id = row.low, i
# Change the trend indicator
up_down = -1
# Current trend is down
elif up_down == -1:
# If got lower than last pivot, update the swing
if row.low < pivot:
# Remove the last pivot, as it wasn't a real one
data.ix[i, 'swings'] = data.ix[last_pivot_id, 'swings'] + (row.low - data.ix[last_pivot_id, 'low'])
data.ix[last_pivot_id, 'swings'] = np.nan
pivot, last_pivot_id = row.low, i
elif row.high > pivot - diff:
data.ix[i, 'swings'] = row.high - pivot
pivot, last_pivot_id = row.high, i
# Change the trend indicator
up_down = 1
print data
Output:
输出：
date close high low open volume swings
2014-05-09 13:30:00 187.56 187.73 187.54 187.70 1922600 NaN
2014-05-09 13:31:00 187.49 187.56 187.42 187.55 534400 NaN
2014-05-09 13:32:00 187.42 187.51 187.35 187.49 224800 -0.35
2014-05-09 13:33:00 187.55 187.58 187.39 187.40 303700 NaN
2014-05-09 13:34:00 187.67 187.67 187.53 187.56 438200 NaN
2014-05-09 13:35:00 187.60 187.71 187.56 187.68 296400 0.36
2014-05-09 13:36:00 187.41 187.67 187.38 187.60 329900 NaN
2014-05-09 13:37:00 187.31 187.44 187.28 187.40 404000 NaN
2014-05-09 13:38:00 187.26 187.37 187.26 187.30 912800 NaN
2014-05-09 13:39:00 187.22 187.28 187.12 187.25 607700 -0.59
回答by tw0000
I updated the answer from @Pawel-Kozela to be compatible with the latest version of pandas, and added an easy way to pass column names.
我更新了@Pawel-Kozela 的答案以与最新版本的Pandas兼容，并添加了一种传递列名的简单方法。
def get_pivots(df, cols=['O','H','L', 'C']):
df['swings'] = np.nan
df.loc[df.index[0], 'swings'] = df.loc[df.index[0], cols[0]]
df.loc[df.index[-1], 'swings'] = df.loc[df.index[-1], cols[0]]
pivot = df.loc[df.index[0], cols[0]]
df.loc[df.index[0], ]
last_pivot_id = 0
up_down = 0
diff = .3
for i, row in df.iterrows():
# We don't have a trend yet
if up_down == 0:
if row[cols[2]] < pivot - diff:
df.loc[i, 'swings'] = row[cols[2]] - pivot
pivot, last_pivot_id = row[cols[2]], i
up_down = -1
elif row[cols[1]] > pivot + diff:
df.loc[i, 'swings'] = row[cols[1]] - pivot
pivot, last_pivot_id = row[cols[1]], i
up_down = 1
# Current trend is up
elif up_down == 1:
# If got higher than last pivot, update the swing
if row[cols[1]] > pivot:
# Remove the last pivot, as it wasn't a real one
df.loc[i, 'swings'] = df.loc[i, 'swings']
df.loc[last_pivot_id, 'swings'] = np.nan
pivot, last_pivot_id = row[cols[1]], i
elif row[cols[2]] < pivot - diff:
df.loc[i, 'swings'] = row[cols[2]] - pivot
pivot, last_pivot_id = row[cols[2]], i
# Change the trend indicator
up_down = -1
回答by Wessel dR
Updated the code of tw0000 as he had a little bug on the lines with 'O' instead of cols[0]
更新了 tw0000 的代码，因为他在用 'O' 而不是 cols[0] 的行上有一个小错误
def get_pivots(df, cols=['O','H','L', 'C']):
df['swings'] = np.nan
df.loc[df.index[0], 'swings'] = df.loc[df.index[0], cols[0]]
df.loc[df.index[-1], 'swings'] = df.loc[df.index[-1], cols[0]]
pivot = df.loc[df.index[0], cols[0]]
df.loc[df.index[0], ]
last_pivot_id = 0
up_down = 0
diff = .3
for i, row in df.iterrows():
# We don't have a trend yet
if up_down == 0:
if row[cols[2]] < pivot - diff:
df.loc[i, 'swings'] = row[cols[2]] - pivot
pivot, last_pivot_id = row[cols[2]], i
up_down = -1
elif row[cols[1]] > pivot + diff:
df.loc[i, 'swings'] = row[cols[1]] - pivot
pivot, last_pivot_id = row[cols[1]], i
up_down = 1
# Current trend is up
elif up_down == 1:
# If got higher than last pivot, update the swing
if row[cols[1]] > pivot:
# Remove the last pivot, as it wasn't a real one
df.loc[i, 'swings'] = df.loc[i, 'swings']
df.loc[last_pivot_id, 'swings'] = np.nan
pivot, last_pivot_id = row[cols[1]], i
elif row[cols[2]] < pivot - diff:
df.loc[i, 'swings'] = row[cols[2]] - pivot
pivot, last_pivot_id = row[cols[2]], i
# Change the trend indicator
up_down = -1
回答by acushner
how about, assuming you only care about highs for the moment:
怎么样，假设你现在只关心高点：
startPx = df.open.iloc[0]
level = ((df.high - startPx) / .3).astype(int)
df['swings'] = level - level.shift(1)
now, to find out what the differences are, you would just do something like:
现在，要找出不同之处，您只需执行以下操作：
changes = df[df.swings != 0]
diffs = changes.high - changes.open.shift(1)
回答by acushner
so i haven't tested this, but something like this will get you what you want. what happens if both the low < pivot - diffand high > pivot + diffin the same minute?
所以我还没有测试过这个，但是这样的事情会让你得到你想要的。如果low < pivot - diff和high > pivot + diff在同一分钟内会发生什么？
def f(df):
pivot = df.open.iloc[0]
diff = .3
def proc(ser):
res = np.nan
if ser.low < pivot - diff:
res, pivot = ser.low - pivot, ser.low
elif ser.high > pivot + diff:
res, pivot = ser.high - pivot, ser.high
return res
df['swings'] = df.apply(proc, axis=1)
