大家好，我正在從事一個專案，當條件變為真時，我只想列印給定條件的第一個和最后一個值...
from astropy.time import Time
import swisseph as swe
from datetime import timedelta as td, datetime
swe.set_ephe_path('G:\Ephemeris')
swe.set_jpl_file('de440.eph')
start_date = '2021-01-01 00:00:00'
end_date = '2021-12-31 00:00:00'
d1 = datetime.strptime(start_date, '%Y-%m-%d %H:%M:%S')
d2 = datetime.strptime(end_date, '%Y-%m-%d %H:%M:%S')
def get_delta(d1, d2):
delta = d2 - d1
return delta
delta = get_delta(d1,d2)
for i in range(delta.days * 1 1):
x = d1 td(days=i)
jd = Time(x).jd
eph = swe.calc_ut(jd, 2, swe.FLG_SPEED swe.FLG_JPLEPH)[0][3]
if eph < 0:
print(x, eph)
輸出 - -
2021-01-31 00:00:00 -0.06355796829817034 <--- Condition true print this value (first value)
2021-02-01 00:00:00 -0.2516828089143771
2021-02-02 00:00:00 -0.4368855305400459
2021-02-03 00:00:00 -0.6135683390488923
2021-02-04 00:00:00 -0.7757942785174291
2021-02-05 00:00:00 -0.9177040258548256
2021-02-06 00:00:00 -1.0340083318420783
2021-02-07 00:00:00 -1.1204908143097292
2021-02-08 00:00:00 -1.1744351937165156
2021-02-09 00:00:00 -1.1948938876715327
2021-02-10 00:00:00 -1.1827399614722631
2021-02-11 00:00:00 -1.140494580853247
2021-02-12 00:00:00 -1.0719696672227224
2021-02-13 00:00:00 -0.9818037732776487
2021-02-14 00:00:00 -0.8749793418345923
2021-02-15 00:00:00 -0.7563965596976655
2021-02-16 00:00:00 -0.6305506630304736
2021-02-17 00:00:00 -0.5013285505901539
2021-02-18 00:00:00 -0.37191597325737136
2021-02-19 00:00:00 -0.24479140317840062
2021-02-20 00:00:00 -0.12177911524216488
2021-02-21 00:00:00 -0.00413500087411478 Print this value (last value)
2021-05-30 00:00:00 -0.004547066831108078 Condition True Print this value (first value)
2021-05-31 00:00:00 -0.07962163144451284
2021-06-01 00:00:00 -0.15215400336574358
2021-06-02 00:00:00 -0.22137358654875092
2021-06-03 00:00:00 -0.28645563729756063
2021-06-04 00:00:00 -0.3465404682847117
2021-06-05 00:00:00 -0.4007583422689736
2021-06-06 00:00:00 -0.44825937136678057
2021-06-07 00:00:00 -0.48824672785691836
2021-06-08 00:00:00 -0.5200109481659706
2021-06-09 00:00:00 -0.5429628624480792
2021-06-10 00:00:00 -0.5566627097783102
2021-06-11 00:00:00 -0.5608427157269316
2021-06-12 00:00:00 -0.5554211491254732
2021-06-13 00:00:00 -0.5405060450116398
2021-06-14 00:00:00 -0.5163887400431304
2021-06-15 00:00:00 -0.4835277090835391
2021-06-16 00:00:00 -0.44252455305021243
2021-06-17 00:00:00 -0.394094049872733
2021-06-18 00:00:00 -0.3390307834305357
2021-06-19 00:00:00 -0.2781753626714067
2021-06-20 00:00:00 -0.21238292731356637
2021-06-21 00:00:00 -0.1424958524480665
2021-06-22 00:00:00 -0.06932129656902836 Print this value (last value)
2021-09-28 00:00:00 -0.08992145079448069 Condition True Print this value (first value)
2021-09-29 00:00:00 -0.2095165387227292
2021-09-30 00:00:00 -0.33367472841201334
2021-10-01 00:00:00 -0.460779986485703
2021-10-02 00:00:00 -0.5885901793012869
2021-10-03 00:00:00 -0.7141710520264541
2021-10-04 00:00:00 -0.8338797124202194
2021-10-05 00:00:00 -0.9434261210721775
2021-10-06 00:00:00 -1.038038248150094
2021-10-07 00:00:00 -1.112746224561704
2021-10-08 00:00:00 -1.162778289034751
2021-10-09 00:00:00 -1.1840315451748653
2021-10-10 00:00:00 -1.1735509206575194
2021-10-11 00:00:00 -1.1299295237178415
2021-10-12 00:00:00 -1.0535472796478889
2021-10-13 00:00:00 -0.9465923196535948
2021-10-14 00:00:00 -0.812857555718251
2021-10-15 00:00:00 -0.6573542581978807
2021-10-16 00:00:00 -0.48582085619604937
2021-10-17 00:00:00 -0.30421497977143097
2021-10-18 00:00:00 -0.11826392085264098 Print this value (last value)
我不知道如何列印這些值，請幫助我我被困在這里，我是絕對的初學者，所以請幫助我解決這個問題......非常感謝！
uj5u.com熱心網友回復：
一種選擇是將每個匹配項 psh 到一個串列中，然后只列印串列中的第一個和最后一個值：
results = []
for i in range(delta.days * 1 1):
x = d1 td(days=i)
jd = Time(x).jd
eph = swe.calc_ut(jd, 2, swe.FLG_SPEED swe.FLG_JPLEPH)[0][3]
if eph < 0:
results.append((x, eph))
print(results[0], results[-1])
uj5u.com熱心網友回復：
您的示例代碼中有太多依賴項，我無法提供經過測驗的解決方案（即不是最少的代碼），但是，作為一般方法，您可以跟蹤先前的值并在滿足條件時列印當前值和先前值（此外系統地列印第一個和最后一個）：
previous = None
for i in range(delta.days * 1 1):
x = d1 td(days=i)
jd = Time(x).jd
eph = swe.calc_ut(jd, 2, swe.FLG_SPEED swe.FLG_JPLEPH)[0][3]
if eph < 0 or previous is None:
if previous: print(*previous)
print(x, eph)
previous = (x, eph)
print(*previous)
uj5u.com熱心網友回復：
與match類似的解決方案，但只保存第一個和最后一個值：
rFirst = None
rLast = None
for i in range(delta.days * 1 1):
x = d1 td(days=i)
jd = Time(x).jd
eph = swe.calc_ut(jd, 2, swe.FLG_SPEED swe.FLG_JPLEPH)[0][3]
if eph < 0:
if rFirst is None:
rFist = (x, eph)
else:
rLast = (x, eph)
print(rFirst, rLast)
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