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如何循环使用多个值who';它在python中的另一个循环之外_Python_Loops - Fatal编程技术网

如何循环使用多个值who';它在python中的另一个循环之外

如何循环使用多个值who';它在python中的另一个循环之外,python,loops,Python,Loops,我在做一个循环,这样我可以迭代一些值: for eachValueOfIntervalQ in range(int(round(qInterval[countQInterval])), x[countQInterval]): print(countQInterval, "1") if filteredFile[eachValueOfIntervalQ] < minValueQ: minValueQ = filteredFile[eachValueOfInte

我在做一个循环,这样我可以迭代一些值:

for eachValueOfIntervalQ in range(int(round(qInterval[countQInterval])), x[countQInterval]):
    print(countQInterval, "1")
    if filteredFile[eachValueOfIntervalQ] < minValueQ:
        minValueQ = filteredFile[eachValueOfIntervalQ]
    countQInterval += 1
allQ.append(minValueQ)
countQInterval = 0
minValueQ = 4000
我希望它遍历所有这些值,以找到与x的间隔相对应的y的最小值,例如从400539

我试过这个

for allIntervalsWhereQCouldBe in x:
    for eachValueOfIntervalQ in range(int(round(qInterval[countQInterval])), x[countQInterval]):
        print(countQInterval, "1")
        if filteredFile[eachValueOfIntervalQ] < minValueQ:
            minValueQ = filteredFile[eachValueOfIntervalQ]
        countQInterval += 1
        allQ.append(minValueQ)
        countQInterval = 0
        minValueQ = 4000
编辑2

rInterval:

[842, 824, 867, 892, 872, 802, 775, 811, 873, 904, 953, 983, 958, 963, 954, 877, 856, 847, 832, 880, 891, 876, 937, 979, 902, 926, 934, 901, 945, 943, 959, 969, 913, 954, 970, 918, 963, 937, 915, 959, 941, 985, 998, 970, 1009, 956, 999, 1011, 970, 1009, 909, 902, 874, 854, 854, 862, 884, 858, 867, 890, 851, 875, 909, 865, 902, 934, 931, 955, 911, 951, 960, 920, 946, 916, 947, 989, 928, 1014, 1026, 978, 952, 961, 919, 988, 1004, 979, 960, 1025, 953, 982, 919, 909, 946, 952, 915, 930, 884, 880, 922, 977, 904, 919, 896, 896, 956, 949, 910, 942, 938, 955, 910, 894, 835, 903, 951, 950, 994, 1025, 962, 945, 857, 851, 870, 863, 927, 904]

我认为您的变量名
allValuesQ
有误导性,当您迭代
x
时,这些将是单个值,对吗?是的,我指的是Q可能存在的所有间隔。但你是对的,必须重写该变量。完成,更新它。如此有效,你只想在
x
中找到每个间隔的
y
的最小值。检查缩进的第一部分,与代码的第二部分相比。最后3行是不同的。这里有太多的值和变量,无法理解您试图实现的目标。我会删除这个问题,并重新发布一个新问题,这是你试图实现的最低限度的代表。简化它。我们不知道什么是
filteredFile
,也不知道其他一些变量。
for rPeaks in range(0, len(x)):
    if rPeaks < (len(x) -1):
        rInterval.append(x[rPeaks +1] - x[rPeaks])

portionOfRInterval = rInterval[0] / 1.2
qAndSInterval = (rInterval[0] - portionOfRInterval)

for eachValueOfX in x:
    qInterval.append(eachValueOfX - qAndSInterval)
    sInterval.append(eachValueOfX + qAndSInterval)
[842, 824, 867, 892, 872, 802, 775, 811, 873, 904, 953, 983, 958, 963, 954, 877, 856, 847, 832, 880, 891, 876, 937, 979, 902, 926, 934, 901, 945, 943, 959, 969, 913, 954, 970, 918, 963, 937, 915, 959, 941, 985, 998, 970, 1009, 956, 999, 1011, 970, 1009, 909, 902, 874, 854, 854, 862, 884, 858, 867, 890, 851, 875, 909, 865, 902, 934, 931, 955, 911, 951, 960, 920, 946, 916, 947, 989, 928, 1014, 1026, 978, 952, 961, 919, 988, 1004, 979, 960, 1025, 953, 982, 919, 909, 946, 952, 915, 930, 884, 880, 922, 977, 904, 919, 896, 896, 956, 949, 910, 942, 938, 955, 910, 894, 835, 903, 951, 950, 994, 1025, 962, 945, 857, 851, 870, 863, 927, 904]